Question: Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}=\frac{1}{5}$ . This triangle is inscribed in rectangle $AQRS$ with $B$ on $\overline{QR}$ and $C$ on $\overline{RS}$ . Find the maximum possible area of $AQRS$ .

设矩形 \(AQRS\) 中

\[
A=(0,0),\quad Q=(w,0),\quad S=(0,h),\quad R=(w,h),
\]

则矩形面积为 \(wh\)。

因为 \(B\) 在 \(QR\) 上、\(C\) 在 \(RS\) 上，可设

\[
B=(w,y),\qquad C=(x,h).
\]

令向量 \(\overrightarrow{AB}\) 与水平边的夹角为 \(\theta\)，向量 \(\overrightarrow{AC}\) 与水平边的夹角为 \(\phi\)。于是

\[
w=40\cos\theta,\qquad x=31\cos\phi.
\]

所以矩形面积

\[
[ AQRS ]=wh=40\cdot31\cos\theta\cos\phi
=1240\cos\theta\cos\phi.
\]

由于 \(B\) 在 \(QR\) 上，其纵坐标不超过 \(h\)，即

\[
40\sin\theta\le 31\sin\phi.
\]

另一方面，给定 \(\sin A=\frac15\)。为了使矩形面积尽可能大，应有 \(\phi>\theta\)，令

\[
\phi-\theta=\alpha,\qquad \sin\alpha=\frac15.
\]

其中 \(\alpha\) 为锐角。

面积变为

\[
1240\cos\theta\cos(\theta+\alpha).
\]

当 \(\theta\ge 0\) 时，

\[
\cos\theta\cos(\theta+\alpha)
\]

在 \(\theta=0\) 时取得最大值。因此

\[
\theta=0,\qquad \phi=\alpha.
\]

此时

\[
w=40,\qquad h=31\sin\alpha=\frac{31}{5}.
\]

还需验证 \(C\) 位于 \(RS\) 上：

\[
x=31\cos\alpha
=31\sqrt{1-\frac1{25}}
=\frac{62\sqrt6}{5}<40,
\]

所以该配置确实可行。

故最大矩形面积为

\[
40\cdot\frac{31}{5}=248.
\]

Final Answer: 248

Explanation:
We need solve Chinese stepwise. Need geometry. Set rectangle A at origin axes, Q=(w,0), S=(0,h), R=(w,h). B on QR x=w, y between; C on RS y=h. Let B=(w,u), C=(v,h). AB=40: w^2+u^2=1600. AC=31 v²+h²=961. sin angle between vectors = |cross|/(1240)=1/5 => |w h - u v|=248. Need maximize wh likely rectangle area. constraints u,v >=0 <= h,w; existence. Could brute parameter optimize wh with dot/cross, side lengths. Scale.

Let normalized coordinates b unit: (cos p,sin p); c unit=(cos q,sin q). Thus w=40 cos p,u=40 sin p; v=31 cos q,h=31 sin q. Also B y<=h:40 sin p <=31 sin q; C x<=w:31 cos q <=40 cos p. sin angle = sin|q-p|=1/5, and cross magnitude. Since q likely >p or less. area wh=1240 cos p cos q. Need constraints. To max cos p cos q, choose q-p = arcsin .2 likely q>p, and p,q around maybe p as low? But C constraint v<=w: 31 cos q <=40 cos p, automatic perhaps depending. B y<=h:40 sin p<=31 sin q implies p lower than q. Thus q>p naturally. Optimize q=p+alpha, alpha=arcsin .2 acute or pi-alpha (bad product/sign). p>=0 and B constraint. Max cos p cos(p+alpha). Different unconstrained max p=-alpha/2, but p>=0 and B constraint demands p maybe? At p=0, B y=0 <= h, yes. product decreases with p near 0 (q positive), so max p=0 perhaps constraints. Then w=40, h=31 sin alpha=6.2, v=31 cos alpha =? sqrt(.96)*31=30.38 <=40. Area 248. Is there possibility q<p and maximize? q=p-alpha. B constraint 40 sin p <=31 sin q likely impossible for positive q since p>q and factor ratio: 40 sinp >31 sinq. Thus no unless q=0 then p=alpha impossible (40 sin alpha >0). Could B possibly y = h boundary? inequality indeed.

Wait inscribed perhaps vertices can lie extensions? B on QR segment yes.

At p=0 configuration A=Q? Q=(w,0), B=(40,0), so B=Q, allowed "on". C upper side. Area 248. Could there be alternate orientation assigning rectangle sides? General as above WLOG coordinates yes A corner and Q,S adjacent.

Could max perhaps cross equation fixed 248 and area wh maybe unbounded? coordinate fixes.

Explain.

Need ensure angle alpha might obtuse with sin .2. q=p+alpha obtuse potentially cos q negative invalid x coordinate, so q <= pi/2; alpha acute only for q>p. If alpha = pi-arcsin, q>90 unless p=0 gives v negative. invalid. Good.

Prove derivative / product.

At p=0 equality boundary, C x=31 cos alpha ≈30.38 <=40. h=31/5.

Chinese step by step.

设矩形的边 \(AQ,AS\) 分别沿 \(x,y\) 轴，且
\[
B=(x,b),\qquad C=(c,y),
\]
其中矩形面积为 \(xy\)。由于 \(B\) 在 \(QR\) 上、\(C\) 在 \(RS\) 上，必须有
\[
x\ge c,\qquad y\ge b.
\]

令从 \(AB\) 到 \(AC\) 的有向角为 \(\delta\)。因为 \(\sin\delta=\frac15\)，而为了使 \(C\) 在 \(B\) 上方，只能取锐角情形：
\[
\cos\delta=\frac{2\sqrt6}{5}.
\]

设 \(AB\) 与 \(x\) 轴夹角为 \(\theta\)，则
\[
B=(40\cos\theta,40\sin\theta),
\]
\[
C=(31\cos(\theta+\delta),31\sin(\theta+\delta)).
\]
条件 \(y\ge b\) 给出
\[
31\sin(\theta+\delta)\ge40\sin\theta.
\]
在该可行范围内，矩形面积
\[
xy=1240\cos\theta\sin(\theta+\delta)
\]
随 \(\theta\) 增大而增大，因此最大面积在等号成立时取得，即 \(B,C\) 有相同的 \(y\)-坐标。

于是写成
\[
B=(x,y),\qquad C=(c,y).
\]
由边长条件，
\[
x^2+y^2=1600,\qquad c^2+y^2=961.
\]
两式相减得
\[
(x-c)(x+c)=639. \tag{1}
\]

另一方面，由向量叉积，
\[
|AB\times AC|=AB\cdot AC\sin A
=40\cdot31\cdot\frac15=248.
\]
但
\[
AB\times AC=xy-yc=y(x-c),
\]
所以
\[
y(x-c)=248. \tag{2}
\]

由（1）（2）可得
\[
x+c=\frac{639}{y},\qquad x-c=\frac{248}{y},
\]
因此
\[
2x=\frac{887}{y},
\]
从而矩形面积为
\[
xy=\frac{887}{2}.
\]

Final Answer: \(\frac{887}{2}\)

Explanation:
We need derive. Coordinate rectangle with A one vertex, rays AS, AQ perpendicular. Need positions projections? Let axes u along AQ (length width), v along AS. B on QR means u-coordinate = x, v between; C on RS means v-coordinate=y, u between. Thus B=(x,b), C=(c,y), rectangle xy. Constraints distances and sin angle.

Need maximize xy where x,y = max coordinates respectively, and existence rectangle axis aligned containing B,C with each assigned opposite side. Equiv x>=c, y>=b, AB²=x²+b²=1600, AC²=c²+y²=961, angle between vectors has sin 1/5 and dot sign ambiguity? Interior A angle perhaps angle between vectors, sin 1/5, dot ± (sqrt24/5)* lengths. Need x y maximize, with b choices and c choices under x>=c y>=b.

Could optimize via rotate coordinates angular. Let B radius40 angle theta; C radius31 angle phi, |sin diff|=.2. Rectangle extents x=max(Bx,Cx), y=max(By,Cy), with B on vertical means Bx>=Cx; C on horizontal Cy>=By. Thus Bx=cx max, Cy=max. So x=40 cos theta likely positive, y=31 sin phi. Conditions coordinates and assignment.

Product =1240 cosθ sinφ. Constraints Cx<=Bx and By<=Cy. maximize.

Angle difference δ with sin=.2. θ, φ likely first quadrant and φ>θ for C above B, giving cos(φ-θ)=±sqrt24/5.

Represent φ=θ+δ. maximize cosθ sin(θ+δ), subject 31 cosφ <=40 cosθ; 40 sinθ <=31 sinφ.

For δ acute: tanδ=1/sqrt24. Product unconstrained max θ=(90-δ)/2, φ=(90+δ)/2. Constraints?
C x/Bx =31 cosφ/(40 cosθ), likely max at angles? φ>θ, ratio <31/40 <1 yes. y condition 40 sinθ <=31 sinφ. Evaluate near 45: 28.28 vs ~31*.7 yes. feasible.
δ obtuse (cos negative): φ-θ=π-arcsin .2, sum near180. Unconstrained same angles sum π/2 still, but θ negative, φ >90: cosθ positive, sinφ positive, coordinates yes. Constraints Cx <= Bx? ratio 31 cosφ negative / positive yes. y ratio 40 sinθ maybe negative <=. Product cosθ sinφ formula perhaps maximum for θ range needing B y >=0 and C x >=0? Wait points could outside first quadrant while rectangle contains A and sides? Since B on QR segment requires y coordinate between 0,y, so B_y>=0. C_x>=0. Thus θ,φ first quadrant actually. For obtuse difference cannot both first quadrant because difference ~168.5 impossible except endpoints no. Thus acute δ. Also reversed φ<θ conflicts Bx>=Cx maybe ratio? B angle higher means By>Cy, violates C y=max. So fixed.

Compute max area. cosθ sinφ with θ+φ=π/2, so sinφ=cosθ, area=1240 cos²θ. δ=φ-θ. θ=(π/2-δ)/2=π/4-δ/2. cos²θ=(1+cos2θ)/2=(1+sinδ)/2 = (1+1/5)/2=3/5. area=744.

Need ensure perhaps rectangle labeling A,Q,R,S sequence: A adjacent Q,S, yes.

Provide stepwise Chinese. Mention feasibility at optimum via inequalities perhaps calculate bounds. At optimum θ ~39.22? δ 11.54, φ50.77. y condition 40 sin39.2=25.3, 31 sin50.8=24? Wait sin50.77=.774, *31=24.0, condition fails! Ah unconstrained not feasible! Indeed B_y > C_y. So boundary y equal likely.

Important optimize constraints.

Conditions:
31 cosφ <=40 cosθ automatically perhaps.
40 sinθ <=31 sinφ.
φ=θ+δ.
At θ lower to satisfy. boundary 40 sinθ=31 sin(θ+δ)=common y. Then product x*y=40 cosθ *40 sinθ=800 sin2θ, maximize θ subject boundary and perhaps θ. Boundary equation. Could potentially boundary gives unique θ? solve. For feasible θ <= root. product increasing until45; root likely ~? 40sinθ=31 sin(θ+δ), root around36? Thus optimum boundary.

Need calculate area at intersection and also Cx<=Bx.

Solve 40 sinθ =31(sinθ cosδ+cosθ sinδ), cosδ=2√6/5? sqrt24/5=2sqrt6/5. sinδ=1/5.
(40-62√6/5) sinθ=31/5 cosθ
tanθ=31/(200-62√6). Rationalize maybe.

Area=800 sin2θ =1600 t/(1+t²). Need exact perhaps simplifies. Alternative geometry with B=(x,y), C=(c,y), BC horizontal and angle sine. Difference vectors B=(40 angle), C=31 angle, same y.

Let x²+y²=1600, c²+y²=961, cross magnitude |x y - y c|=y(x-c)= (1/5)*1240=248. Also x²-c²=639=(x-c)(x+c). Let d=x-c=248/y, s=x+c=639/y. x=(887)/(2y), c=391/(2y). Then y²+c²=961: y² + 391²/(4y²)=961. Let z=y². 4z²-3844z+152881=0. discriminant 3844²-4*152881. Calculate 14,776,? 3844²=(3800+44)^2=14,440,000+334400+1936=14,776,336. minus611524=14,164,812. sqrt? factor. approximate sqrt3763.6. z=(3844 ± sqrt)/8. Two solutions y ~31 or ~14.8? If y=31 then c=0, cross=248 requires d=8 x? x ~? sqrt639=25.28 not. exact roots.

Area xy=x*y=(887)/(2)=443.5 invariant! Interesting each same-y configuration area 887/2. Wait x y=887/2. Both branches, feasible c positive. So answer 887/2, not 744. Check assignment x>=c yes. Great.

But earlier cross signed dot angle could vectors orientation C angle > B, cross x*y_C - y_B*c = y(x-c) positive fixed.

Maximum maybe configurations not same y with product lower, boundary maximum argument. Need present proof via optimization. δ acute, feasible θ upper root and maybe lower constraints Cx. Area f=1240 cosθ sin(θ+δ). On feasible θ from 0 to root (<45), increasing, so boundary. State root and monotonic.

Final numeric perhaps 887/2. User says 数值.

设矩形 \(AQRS\) 的顶点依次为
\[
A=(0,0),\quad Q=(w,0),\quad R=(w,h),\quad S=(0,h),
\]
则矩形面积为 \(wh\)。

令射线 \(AB\) 与 \(AQ\) 的夹角为 \(\beta\)，射线 \(AC\) 与 \(AQ\) 的夹角为 \(\gamma\)。因为 \(B\) 在 \(\overline{QR}\)、\(C\) 在 \(\overline{RS}\)，所以
\[
0\le \beta\le \gamma\le \frac{\pi}{2}.
\]

由 \(AB=40,\ AC=31\)，两向量的夹角为
\[
\gamma-\beta.
\]
利用两向量叉积公式，
\[
AB\cdot AC\cdot \sin A=|\overrightarrow{AB}\times\overrightarrow{AC}|.
\]
而
\[
|\overrightarrow{AB}\times\overrightarrow{AC}|
=40\cdot31\sin(\gamma-\beta).
\]
已知 \(\sin A=\frac15\)，故
\[
40\cdot31\sin(\gamma-\beta)
=40\cdot31\cdot\frac15,
\]
所以
\[
\sin(\gamma-\beta)=\frac15.
\]

矩形面积为
\[
wh=(40\cos\beta)(31\sin\gamma)
=1240\cos\beta\sin\gamma.
\]
记 \(\delta=\gamma-\beta\)，则 \(\sin\delta=\frac15\)，且 \(0<\delta<\frac{\pi}{2}\)。于是
\[
[\![AQRS]\!]
=1240\cos\beta\sin(\beta+\delta).
\]
利用积化和差，
\[
\cos\beta\sin(\beta+\delta)
=\frac12\bigl(\sin(2\beta+\delta)-\sin\delta\bigr).
\]
当
\[
2\beta+\delta=\frac{\pi}{2}
\]
时，上式取得最大值。因此
\[
\max \cos\beta\sin(\beta+\delta)
=\frac12\left(1-\frac15\right)=\frac25.
\]
所以矩形最大面积为
\[
1240\cdot\frac25=496.
\]

Final Answer: 496

Explanation:
We need derive. Need understand rectangle, A vertex, triangle inscribed B,C on two non-A adjacent? Rectangle AQRS cyclic labels, QR opposite AQ, RS opposite AQ? Coordinates A origin rectangle sides axes, Q=(w,0), R=(w,h), S=(0,h). B on QR x=w, C on RS y=h. AB=40 AC=31. Let B=(w,y), C=(x,h). Dot constraints.

AB: w^2+y^2=1600. AC x^2+h^2=961. Angle A between AB and AC sin=1/5. Thus |cross|/(1240)=1/5 => |wh-xy|=248. Rectangle area wh maximize subject 0<=y<=h, 0<=x<=w, and lengths.

Need optimize. Parametrize B angle beta: w=40 cos beta, y=40 sin beta, beta in [0,pi/2]. C angle gamma: x=31 cos gamma,h=31 sin gamma gamma [0,pi/2]. Inclusions impose w<=x? B y<=h and C x<=w. But likely rectangle boundary rays order C angle > beta, so gamma>beta, inclusion.

Area = wh=1240 cos beta sin gamma. Cross abs=1240 sin(gamma-beta)=248 => sin(delta)=1/5. gamma-beta delta either acute arcsin 1/5 or pi - arcsin, impossible within pi/2 maybe obtuse possible. Also perhaps orientation reversed gamma<beta, but boundary inclusion C x<=w implies cos gamma <= cos beta => gamma>=beta. So delta in [0,pi/2], acute only. Thus gamma=beta+delta, delta fixed acute.

Maximize cos beta sin(beta+delta), beta >=0, gamma<=pi/2. Derivative: -sin beta sin(...) + cos beta cos(...)=cos(2beta+delta)=0 => 2beta+delta=pi/2. beta=(pi/2-delta)/2=pi/4-delta/2. Area compute 1240 cos beta sin(beta+delta). product = 1240/2 [sin(delta)+sin(2beta+delta)] =620[1/5+1]=744.

Check endpoint maybe lower. So 744.

But potential "inscribed" can B on lines/segments and orientation perhaps cross negative incompatible inclusions? Need articulate stepwise Chinese. Also concern rectangle labels maybe AQRS means A-Q-R-S, yes.

Can derive coordinates. Need show delta fixed and interval. inclusion implies angles order.

At optimum beta ~39.3 deg, gamma ~50.7, dimensions etc valid. area.

Chinese response step by step, final exact mandated last line. Ensure only numeric after.

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Question: Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes.

Let
\[
N(m)=m^2-m+11.
\]

### 1. Rule out small prime factors

Since \(m(m-1)\) is even, \(N(m)\) is odd.

Also, checking residues gives
\[
N(m)\not\equiv 0\pmod 3,\qquad
N(m)\not\equiv 0\pmod 5,\qquad
N(m)\not\equiv 0\pmod 7.
\]
Therefore every prime factor of \(N(m)\) is at least \(11\).

If \(N(m)\) is a product of at least four primes, then
\[
N(m)\ge 11^4=14641.
\]
Now
\[
N(121)=121^2-121+11=14531<14641,
\]
and \(N(m)\) is increasing for positive \(m\). Hence
\[
m\ge 122.
\]

### 2. Check the range below \(132\)

Suppose \(122\le m\le131\). Then
\[
N(m)\le N(131)=17041.
\]
If \(N(m)\) has at least four prime factors, its smallest prime factor \(p\) must satisfy
\[
p^4\le 17041.
\]
Since \(p\ge11\) and \(13^4=28561>17041\), we must have \(p=11\). Thus \(11\mid N(m)\).

But
\[
N(m)\equiv m^2-m=m(m-1)\pmod{11},
\]
so \(11\mid N(m)\) exactly when \(m\equiv0\) or \(1\pmod{11}\). Among \(122,\ldots,131\), this happens only for \(m=122\).

However,
\[
N(122)=14773=11\cdot17\cdot79,
\]
which has only three prime factors. Therefore no \(m<132\) works.

### 3. Verify \(m=132\)

\[
N(132)=132^2-132+11=17303=11^3\cdot13.
\]
This is a product of four primes, counting multiplicity.

Final Answer: 132

Explanation:
We need solve. Need reason step-by-step, final exact. Need understand "product of at least four not necessarily distinct primes" means integer has prime factorization total multiplicity >=4, positive likely. Need least positive m such n=m^2-m+11 has Ω(n)>=4. Need determine.

Need carefully search/prove minimality. Polynomial values. Small m:

m=1:11 prime Ω1
2:13
3:17
4:23
5:31
6:41
7:53
8:67
9:83
10:101
11:121=11^2 Ω2
12:143=11*13 Ω2
13:167 prime?
14:193
15:221=13*17
16:251 prime?
17:283 prime?
18:317
19:353
20:391=17*23?
21:431 prime?
22:473=11*43?
23:517=11*47
24:563?
25:611=13*47?
26:661?
27:713=23*31
28:767=13*59
29:823?
30:881?
Maybe first four-factor much later.

We need find least m. Need devise systematic math. Polynomial discriminant -43. Values likely often prime for m up to some Euler polynomial-ish. m^2-m+11 related Heegner 43 and m^2-m+11 prime for m=1..10 only; at 11 composite. At least 4 may perhaps m=32? Need search intelligently.

Factorization conditions. Need ensure no earlier Ω>=4. Could manually enumerate factor values using primes and CRT. Since n grows ~m². Four not necessarily distinct implies divisible by p^? Or product factors; minimum product among 4 primes is 2^4=16 but parity always? m²-m even, +11 odd, so no 2. Minimum odd 3^4=81. Need characterize Ω>=4. Need search m upward and establish all below don't.

Could derive via divisors and factorization table. Need compute.

Let's inspect sequence OEIS perhaps polynomial x^2-x+11. Find first value with four prime factors. We can enumerate.

N(m)=m(m-1)+11, odd. Divisibility by p means m(m-1) ≡ -11 mod p. This has roots if discriminant 1-44=-43 is square mod p. For p=11? discriminant -43 ≡? -43 mod11=1, roots [1±1]/2 =0,1 mod11, factors valuations potentially p² due derivative 2m-1 not 0 at roots mod11 (derivative ±1), so exactly one root each? At m≡0 or1, divisible by 11, could powers via lifts but value modulo121. p=43? discriminant 0, root m=22 mod43; derivative ≡0, special. Other primes with Kronecker (-43/p)=1.

To get Ω>=4. Could search m and factor.

Maybe four-factor first around 80. Let's calculate sequence factors perhaps leverage q divisibility.

Question likely contest problem intended clever. Polynomial N = m²-m+11 = (m-5)(m+4)+31? Also Euler-like, known first with 4 factors perhaps m=32: 32²-32+11=1003=17*59 (2). m=33 1067=11*97? m34 1133=11*103; m35 1201 prime; etc.

At least four prime factors can be as semiprime? Need multi-prime. Maybe m=55? N=2981=11*271 only. m=56 3091=11*281. Powers? At m where divisible 11, quotient (m(m-1))/11+1. Need quotient have Ω>=3. E.g. m=??.

Could be p=43 leading powers. At m=22, N=473=11*43 (Ω2). m=65? N=4171? divisibility roots.

Let's generate systematically.

We need find first m with Ω(N)>=4. Possible patterns:
- N divisible by 3 if m(m-1) ≡2 mod3. Products are 0 or2? m residues: 0->0,1->0,2->2, so m≡2 mod3. Then quotient. Earliest m=2 N13 not 3? 2*1+11=13 wait -11 mod3 =1, equation product ≡1? Let's recalc -11 ≡1 mod3. Product residues 0,0,2 (2≡-1), never 1. So never divisible by3! Indeed discriminant -43 ≡? -1 mod3=2 nonresidue. No factor3.
p=5: -11 ≡4, products mod5 0,0,2,1,2; never4. no 5.
p=7: -11=3; products:0,0,2,6,6,2,0 never3. no7.
p=11 roots yes.
p=13: -11=2; products sequence m(m-1): 0,0,2 yes roots m=2? and? 2*1=2, 3*2=6,4*3=12=12,5*4=20=7,6*5=30=4,7*6=42=2 => roots2,7. So every m≡2,7 mod13.
p=17: -11=6; products residues perhaps roots? discriminant -43 mod17 = -9=8; square roots? 5²=8, roots (1±5)/2: 3,? -4/2=-2=15. yes m3,15.
p=19: -43 mod19=-5=14, squares? 6²=36=17, 8²=64=7, 9²=81=5, 10²=5, 11²=7,13²=17; 14 not square perhaps no.
p=23: -43=3 mod23, squares include? 7²=3, roots (1±7)/2=4, -3=20. yes.
p=29: -43=14, squares? 6²=7,7²=20,8²=6,9²=23,10²=13,11²=5,12²=28,13²=24,14²=22; maybe 16²=24 etc no14.
p=31: -43=19; squares? 9²=19, roots (1±9)/2=5, -4=27. yes.
p=37: -43=31; squares? 8²=27, 9²=7,10²=26,11²=10,12²=33,13²=21,14²=11,15²=3,16²=34,17²=30; no31 perhaps.
p=41: -43=39=-2; squares? 10²=18, etc maybe -2 square since p ≡1 mod8, yes roots sqrt 17? 17²=289=2? 41*7=287 =>2, so -2 =24²? 17²=2, (-17)² 24²=2 not -2. sqrt -1 mod41=9, sqrt2=17, product12²=144=21 no. 9*17=153=30, square=900 mod41=39 yes sqrt30. roots (1±30)/2; inverse2=21:31*21=651 mod41=36? 41*15=615 rem36; (-29)*21=-609 mod41=6. roots 6,36.
p=43 root22.
Thus divisors sparse.

Could search products of four prime factors under sqrt-ish. Since N<m². If Ω>=4 and no tiny allowed primes 3,5,7 (allowed smallest 11), then N>=11^4=14641 unless repeated factors? Ah allowed primes include 11 and maybe? Check 2 no,3,5,7 impossible. Thus m²-m+11 >=14641 -> m roughly122. This is huge! Wait smallest prime allowed could be 11 indeed if no 2,3,5,7. For 4 primes all >=11, minimum 14641. Thus m≥122. That's lower bound. This simplifies. Search m from 122 upward perhaps first near. But N may have factor p=11 and three others, so likely first around 122. Need factor sequence until hit.

At m=122: N=14884? 122²=14884 -122 +11=14773. 11? m≡1 mod11? 122≡1, yes N divisible 11. quotient 1343. Factor 1343: divisibility primes >=11 sqrt36.6. 11: 11*122=1342 rem1; 13:13*103=1339 rem4;17*79=1343 exactly? 17*80=1360-17=1343. So N=11*17*79 Ω3, not 4. Need at least quotient Ω3.

m=123 not div 11 (3?); N=15017? 15129-123+11=15017. Factor maybe. Need enumerate until four factors. Lower bound near 122 but allowed primes include 13 etc.

Could be immediately m? Need find.

Any N with Ω≥4 and m just over 122 must factors all small (11 etc), and N has no prime >? Since four min. We can characterize via divisibility products.

For m≥122, N grows. Search perhaps around m=134 etc. We need exact first. Let's establish allowable prime divisors: primes p such that -43 quadratic residue mod p, excluding 2,3,5,7. First: 11,13,17, then? 19 no, 23 yes, 29 no,31 yes,37 no,41 yes,43 special,47? Let's calculate. For N around 15k, Ω4, factors likely small. Could enumerate candidate integers with four prime factors in sequence using roots.

We can search m sequentially factorization with manageable effort. Starting 122. But could take perhaps many.

Alternative exploit N minimum all allowed factors. If m< maybe 200, N<~40k. Any Ω≥4 has smallest factors and at least? We can enumerate quartets allowed whose product lies in polynomial range and solve roots / factor.

Could identify divisibility by p powers. For total multiplicity 4, N must be divisible by some product of 4 allowed primes ≤N. In particular either:
- p^4 for p=11,13,17,...
- p^3 q
- p²q²
- p² q r
- p q r s.
Need find minimal m ≥122 satisfying any.

A straightforward manual table of N factorization from 122 onward. Let's estimate likely hit. Divisibility probabilities 2/p. For allowed p ~. Ω distribution. Probability Ω≥4 maybe 5-10%, so within ~10-30 values. Could calculate.

Let's factor m=122 onward. Need accurate. Generate N recurrence N(m+1)=N(m)+2m. Starting N122=14773. Then increments 244,246,...

Let's list and factor using divisibility. Could use modular root residues to quickly identify allowed small primes. We need detect Ω>=4 only, but factor values enough.

Allowed primes list up to maybe sqrt N (~200). Determine roots. Manual divisibility tests.

Let's start.

m=122 N14773 = found 11*17*79 Ω3.

m=123: N=14773+244=15017. Need factor. Test p allowed:
11: m 123≡2 no.
13 roots2,7; 123 mod13=6 no.
17 roots3,15; 4 no.
23 roots4,20; 8 no.
31 roots5,27; 30 no.
41 roots6,36; 0 no.
43 root22; 37 no.
47 need determine whether allowed and roots. Could direct test N/47: 47*319=14993, *320=15040, no.
53: 53*283=14999, *284=15052 no.
59:59*254=14986, *255=15045 no.
61:61*246=15006, *247=15067 no.
67:67*224=15008, *225=15075 no.
71:71*211=14981, *212=15052 no.
73:73*205=14965,*206=15038 no.
79:79*190=15010,*191=15089 no.
83:83*181=15023 no.
89:89*169=15041 no.
97:97*155=15035.
101:101*149=15049.
103:103*146=15038.
107:107*140=14980, rem37.
109:109*138=15042.
113:113*133=15029.
127:127*118=14986.
131*115=15065.
137*110=15070.
139*108=15012.
149*101=15049.
sqrt15017≈122.5 actually only primes ≤122! I overshot. So likely prime if none. But need allowed p and we only tested up to113. N prime? Could have disallowed p? Any divisor necessarily roots criterion; list allowed ≤122. We should know all, direct tests enough if missed. We tested all primes regardless from47 onward and no. Below 43 only allowed 11,13,17,23,31,41,43 and no. 19,29,37 impossible. So 15017 prime. Fine.

m=124 N=15017+246=15263. Factor: check small roots:
11 124≡3 no
13 7? 124 mod13=7 yes divisible 13. 15263/13=1174.076? 13*1174=15262, remainder1! Wait root calculation perhaps divisibility root m=7 should N divisible13. Compute N:124²=15376-124+11=15263. Mod13: m7 =>49-7+11=53 ≡1, not 0. Our root derivation error. Solve product m(m-1) ≡ -11 mod13 =2. m=7: 7*6=42 ≡3, not2 (I said2 incorrectly). Let's recompute roots. Discriminant D=1-44=-43 mod13 = -4 =9. sqrt=±3. roots (1±3)/2: 4/2=2; (-2)/2=-1=12. So roots 2,12, not 7. Right. Need correct roots lists maybe others.

Let's systematically establish using quadratic formula carefully.

For p:
D=-43.
p=11 D=1 roots (1±1)/2 =1,0 yes.
p=13 D=9 sqrt3 roots2,12.
p=17 D=-9=8 sqrt? 5²=8 roots (1±5)/2: 6/2=3; -4/2=-2=15 yes.
p=23 D=3 sqrt? 7²=49=3, roots 8/2=4; -6/2=-3=20 yes.
p=31 D=19 sqrt9, roots10/2=5; -8/2=-4=27 yes.
p=41 D=39 sqrt? 30²=900 rem39; roots31*inv2(21):651 rem? 41*15=615 rem36 yes; other (1-30=-29)*21=-609; -609 + 615=6. roots36,6 yes.
p=43 D=0 root1/2=22.
Good.

Then p=47 D=4 sqrt2, roots (1±2)/2: 3/2 =? inv2=24 =>72=25 mod47; other -1/2=23. roots25,23. So allowed.
p=53 D=10; squares? 14²=196 rem37, etc find sqrt10: 13²=169=10! roots (1±13)/2 inv2=27:14*27=378 rem7; -12*27=-324 rem? -324+371=47. roots7,47.
p=59 D=16 sqrt4 roots5/2=3; -3/2=-? 3*30=90=31? Wait inv2=30: 5*30=150 rem32, other -3*30=-90 rem -31=28. roots32,28.
p=61 D=18; sqrt? 17²=289 rem45; 18 maybe 21²=441 rem14; 23²=529 rem41; 25²=625 rem15; 29²=841 rem48. Need determine. D Kronecker perhaps. Could defer.

Let's factor 124 direct: N=15263. Test small:
11 no m mod11=3.
13 no (2,12; 124 mod13=7).
17 roots3,15; 124 mod17=5 no.
23 roots4,20; 9 no.
31 roots5,27; 0 no.
41 roots6,36; 1 no.
43 root22; 38 no.
47 roots23,25; 124 mod47=30 no.
53 roots7,47;18 no.
Could divide perhaps allowed p≥59. sqrt 123.5. Test direct maybe N factors. 15263 /? 59:59*259=15281 no. 61*250=15250 rem13. 67*228=15276. 71*215=15265 rem -2. 73*209=15257 rem6. 79*193=15247 rem16.83*184=15272. 89*171=15219 rem44. 97*157=15229. 101*151=15251. 103*148=15244. 107*143=15301.109*140=15260 rem3.113*135=15255.127 >sqrt. likely prime.

m=125 N=15263+248=15511. Small root:
125 mod11=4 no
mod13=8 no
mod17=6 no
mod23=10
mod31=1? roots5,27 no
mod41=2
mod43=39
mod47=31
mod53=19
Check direct allowed >=59 up to124. Maybe factor. 15511:
59*263=15517
61*254=15494
67*231=15477; *232=15544
71*218=15478;*219=15549
73*212=15476;*213=15549
79*196=15484;*197=15563
83*187=15521
89*174=15486;*175=15575
97*160=15520
101*154=15554
103*151=15553
107*145=15515
109*142=15478;*143=15587
113*137=15481;*138=15594
127 > sqrt (124.5). prime likely.

m=126 N=15511+250=15761. Small:
126 mod11=5 no
13=9 no
17=7
23=11
31=2
41=3
43=40
47=32
53=20
Potential.
15761 sqrt125.5. Check:
59*267=15753 rem8
61*258=15738 rem23
67*235=15745 rem16
71*222=15762 rem-1
73*216=15768
79*199=15721 rem40
83*190=15770
89*177=15753
97*162=15714
101*156=15756
103*153=15759 rem2
107*147=15729
109*145=15805
113*139=15707
127 >. prime likely.

m=127 N=15761+252=16013. Ah 16013 maybe factors. Small:
127 mod11=6 no; 13=10 no;17=8;23=12;31=3;41=4;43=41;47=33;53=21.
Direct:
59*271=15989; *272=16048
61*262=15982;*263=16043
67*239=16013 exactly? 67*240=16080-67=16013 yes! N=67*239, both prime Ω2. (67 allowed likely).
m=128 N=16013+254=16267. m mod11=7 no; mod13=11 no;17=9;23=13;31=4;41=5;43=42;47=34;53=22.
Could factor. 16267 perhaps 7 impossible. sqrt127.5.
59*276=16284
61*267=16287
67*243=16281
71*229=16259 rem8
73*223=16279
79*206=16274
83*196=16268 rem-1
89*183=16287
97*168=16296
101*161=16261 rem6
103*158=16274
107*152=16264 rem3
109*149=16241
113*144=16272
127? >sqrt just. likely prime.

m=129 N=16267+256=16523. m mod11=8 no; mod13=12 yes -> divisible13. 16523/13:13*1271=16523 exactly. Factor1271. sqrt35.6, allowed 11,13,17,23,31. 1271 mod11? 11*115=1265 rem6. /13:13*97=1261 rem10. /17:17*74=1258 rem13. /23:23*55=1265 rem6. /31:31*41=1271 exactly. Thus N=13*31*41 Ω3. Not 4.

m=130 N=16523+258=16781. Small m mod11=9 no; 13=0 no;17=11;23=15;31=6;41=7;43=1;47=36;53=24.
Factor? sqrt129.5.
59*284=16756 rem25
61*275=16775 rem6
67*250=16750 rem31
71*236=16756 rem25
73*230=16790
79*212=16748 rem33
83*202=16766 rem15
89*189=16821
97*173=16781? 97*(170+3)=16490+291=16781 yes. N=97*173 Ω2.

m=131 N=16781+260=17041. Ah 130²? Known 17041 =? 41*? 41*416=17056; perhaps 11? m mod11=10 not root; 13=1 no; 17=12;23=16;31=7;41=8;43=2;47=37;53=25.
sqrt130.5.
59*289=17051
61*279=17019 rem22
67*254=17018
71*240=17040 rem1
73*233=17009
79*216=17064
83*205=17015
89*191=16999
97*176=17072
101*169=17069
103*165=16995
107*159=17013
109*156=17004
113*151=17063
127*134=17018
likely prime.

m=132 N=17041+262=17303. m mod11=0 root. /11=1573 exactly? 1573*11=17303. Factor1573 sqrt39.7: allowed 11,13,17,23,31,37(no).
1573/11=143 exactly? 11*143=1573. Ah so 11²*143 =11³*13. Thus N=11^3*13, total multiplicity 4! m=132. Likely answer, assuming no earlier m≥122. But need rigorously prove least. We found all m122-131 none Ω≥4 via claimed factor statuses. Need present elegant exhaustive reasoning, not list primality tedious perhaps exploit lower bound and divisibility constraints.

At m=132 N=17303=11³*13 yes compute: 132²=17424 -132+11=17303. 1331*13=17303. Great. Need prove no m<132. We used lower bound: N odd and not divisible by3,5,7. Show any prime divisor ≥11. Then if four primes, N≥11^4=14641. Solve m²-m+11≥14641. m=121? N=14531? 121²-121+11=14531 (<14641). m=122 N14773 ≥. Thus m≥122. Then need show for 122≤m≤131 N has at most3 prime factors. Could factor each. We need provide concise table factorization showing Ω. We can factor all ten. Our calculations suggested some prime; need verify each exactly and factorization. We should ensure correctness and present perhaps factorizations for all. Need factor 15017,15263,15511,15761,16267,17041 claimed prime. Need rigorously verify, and factor 16013=67*239,16781=97*173. Others:
122 14773=11*17*79.
123 15017 maybe factor we might have missed: let's factor accurately. sqrt=122.54. We tested many but need systematically check primes allowed. Could be 15017= ? Let's use primality checks all primes. We should verify exact.
124 15263 perhaps composite with factors >? sqrt123.5, we tested most.
125 15511.
126 15761.
128 16267.
131 17041.
Need be absolutely sure. Could find factor via modular arithmetic or actual primality.

Let's independently verify all factorizations and perhaps optimize proof via modular exclusions rather than primality. Table can state factorization with "prime" but need establish primes. For contest answer, enough table perhaps primes can be checked; but should avoid unsupported. Could show smallest prime factors / products.

Let's calculate all accurately using computational-like manual.

First derive allowed prime divisors criterion and table residues perhaps make checking easy. Since N odd, congruent mod primes:
- mod3: m²-m = m(m-1) ≡0 or2; +11 ≡ +2 =>2 or1, never0.
- mod5: products 0,0,2,1,2; +1 (11≡1) gives1,1,3,2,3 no0.
- mod7: products 0,0,2,6,6,2,0; +4 gives4,4,6,3,3,6,4 no0.
So min prime factor 11.

Then Ω≥4 -> N≥14641. For m positive monotonic m≥? N(121)=? 121²=14641! Ah 121²=14641. N=14641-121+11=14531. Yes. N(122)=14773. So m≥122. If m=132 works, only check 122..131.

Could use factor table. Let's verify.

N values recurrence:
m 122: 14773
123: +2*122=244 ->15017
124 +246=15263
125 +248=15511
126 +250=15761
127 +252=16013
128 +254=16267
129 +256=16523
130 +258=16781
131 +260=17041
132 +262=17303.

Factor checks:

14773:
11*1343; 1343=17*79 (17*80 1360-17). yes Ω3.

15017:
Try divisibility all primes <=122. Let's systematically test with modular quotient perhaps factor hidden like 15017= ? Sum digits14 no3; no5; 7: 7*2145=15015 rem2. 11: 11*1365=15015 rem2.13:13*1155=15015 rem2. 17:17*883=15011 rem6. 19:19*790=15010 rem7. 23:23*653=15019 rem-2. 29:29*518=15022 rem-5. 31:31*484=15004 rem13. 37:37*406=15022. 41:41*366=15006 rem11. 43:43*349=15007 rem10. 47*319=14993 rem24. 53*283=14999 rem18. 59*254=14986 rem31. 61*246=15006 rem11. 67*224=15008 rem9. 71*211=14981 rem36. 73*205=14965 rem52. 79*190=15010 rem7. 83*181=15023 rem-6. 89*168=14952 rem65 (earlier I used169=15041). 97*154=14938 rem79 (155=15035). 101*148=14948 rem69 (149=15049). 103*145=14935 rem82 (146=15038). 107*140=14980 rem37. 109*137=14933 rem84 (138=15042). 113*132=14916 rem101 (133=15029). Also prime 2,3,5 etc no. sqrt floor122, primes up to113 indeed. prime.

15263:
Check 7:7*2180=15260 rem3. 11:11*1387=15257 rem6. 13:13*1174=15262 rem1. 17:17*898=15266 rem-3. 19:19*803=15257 rem6. 23:23*663=15249 rem14. 29:29*526=15254 rem9. 31:31*492=15252 rem11. 37:37*412=15244 rem19. 41:41*372=15252 rem11. 43:43*355=15265 rem-2.47:47*325=15275 rem-12.53:53*288=15264 rem-1.59:59*259=15281 rem-18.61:61*250=15250 rem13.67:67*228=15276 rem-13.71:71*215=15265 rem-2.73:73*209=15257 rem6.79:79*193=15247 rem16.83:83*184=15272 rem-9.89:89*171=15219 rem44.97:97*157=15229 rem34.101:101*151=15251 rem12.103:103*148=15244 rem19.107:107*142=15194 rem69; *143=15301.109*140=15260 rem3.113*135=15255 rem8. sqrt 123.5, includes prime 127 no. prime.

15511:
Check maybe 7:7*2216=15512 rem-1. 11:11*1410=15510 rem1. 13:13*1193=15509 rem2.17:17*912=15504 rem7.19:19*816=15504 rem7.23:23*674=15502 rem9.29:29*535=15515 rem-4.31:31*500=15500 rem11.37:37*419=15503 rem8.41:41*378=15498 rem13.43:43*361=15523 rem-12.47:47*330=15510 rem1.53:53*293=15529 rem-18 (292=15476 rem35).59*263=15517 rem-6.61*254=15494 rem17.67*231=15477 rem34.71*218=15478 rem33.73*212=15476 rem35.79*196=15484 rem27.83*187=15521 rem-10.89*174=15486 rem25.97*160=15520 rem-9.101*153=15453 rem58.103*150=15450 rem61.107*145=15515 rem-4.109*142=15478 rem33.113*137=15481 rem30. sqrt124.5 includes 127 no, and prime 2.. Prime.

15761:
7:7*2251=15757 rem4.
11:11*1433=15763 rem-2.
13:13*1212=15756 rem5.
17:17*927=15759 rem2.
19:19*829=15751 rem10.
23:23*685=15755 rem6.
29:29*543=15747 rem14.
31:31*508=15748 rem13.
37:37*426=15762 rem-1.
41:41*384=15744 rem17.
43:43*366=15738 rem23 (367=15781).
47:47*335=15745 rem16.
53:53*297=15741 rem20.
59*267=15753 rem8.
61*258=15738 rem23.
67*235=15745 rem16.
71*222=15762 rem-1.
73*215=15695 rem66;216=15768.
79*199=15721 rem40.
83*189=15687 rem74;190=15770.
89*177=15753 rem8.
97*162=15714 rem47.
101*156=15756 rem5.
103*153=15759 rem2.
107*147=15729 rem32.
109*144=15696 rem65;145=15805.
113*139=15707 rem54.
sqrt125.5 (prime 127 no). prime.

16013=67*239. Check both prime yes 239 prime, 67. Ω2.

16267:
7:7*2324=16268 rem-1
11:11*1479=16269 rem-2
13:13*1251=16263 rem4
17:17*957=16269 rem-2
19:19*856=16264 rem3
23:23*707=16261 rem6
29:29*561=16269 rem-2
31:31*525=16275 rem-8
37:37*439=16243 rem24;440=16280
41:41*397=16277 rem-10
43:43*378=16254 rem13
47:47*346=16262 rem5
53:53*307=16271 rem-4
59*275=16225 rem42;276=16284
61*266=16226 rem41;267=16287
67*242=16214 rem53;243=16281
71*229=16259 rem8
73*222=16206 rem61;223=16279
79*205=16195 rem72;206=16274
83*196=16268 rem-1
89*182=16198 rem69;183=16287
97*167=16199 rem68;168=16296
101*161=16261 rem6
103*158=16274 rem-7
107*152=16264 rem3
109*149=16241 rem26
113*144=16272 rem-5
sqrt127.54, need check prime 127? 127²=16129; sqrt 16267 >127? 127²=16129, 128²=16384, so sqrt between127 and128, includes127. Earlier omitted. 127*128=16256 rem11, so no. Prime.

16523 factor 13*1271; 1271=31*41. So 13*31*41, Ω3. Check multiplication 31*41=1271 yes.

16781=97*173; both prime. Ω2.

17041: Need check sqrt130.54 includes127. We tested many; let's ensure no factor. Maybe known 17041= ? 131? 130²=16900; N=17041. Try 7:7*2434=17038 rem3.
11:11*1549=17039 rem2
13:13*1311=17043 rem-2
17:17*1002=17034 rem7
19:19*897=17043 rem-2
23:23*741=17043 rem-2
29:29*? 29*587=17023 rem18;588=17052
31:31*? 31*550=17050 rem-9
37:37*? 37*460=17020 rem21
41:41*? 41*416=17056 rem-15
43:43*? 43*396=17028 rem13
47*? 47*362=17014 rem27
53*? 53*321=17013 rem28
59*289=17051 rem-10
61*279=17019 rem22
67*254=17018 rem23
71*240=17040 rem1
73*233=17009 rem32
79*215=16985 rem56;216=17064
83*205=17015 rem26
89*191=16999 rem42
97*175=16975 rem66;176=17072
101*168=16968 rem73;169=17069
103*165=16995 rem46
107*159=17013 rem28
109*156=17004 rem37
113*150=16950 rem91;151=17063
127*134=17018 rem23
Prime.

So table fine. But perhaps we can prove no Ω≥4 without claiming primality via lower bounds and known divisors. For N in 14773 to17041. If Ω≥4 and all factors≥11. We can derive constraints based on divisibility by 11 or 13 etc. Maybe make table simpler:

For each m in 122..131, factor at least something:
- m=122,129,132 divisible etc.
Others may be prime/semiprime.
Could show by checking if N has four prime factors, its smallest factor p ≤ N^{1/4} ≈ 11.3–11.5! Ah powerful! Since product of four primes all ≥ p, the smallest prime factor p ≤ fourth root N. For N around 15k–17k, fourth root ~11.3. Since only allowed primes ≥11, p must be 11 (integer prime). Indeed fourth root of 17041: 11^4=14641, 12^4=20736; prime 11. Any Ω≥4 N<20736 must be divisible by 11. Great. For 122≤m≤131, check N divisible by11 only at m≡0 or1 mod11: m=122 (≡1) and? 132 outside; 121 below; among122-131 none except122! Wait m=122 is 1 mod11. So only m=122 candidate. Its factorization Ω3. This avoids factoring all others! Excellent. Need check bound upper: N(131)=17041 < 13^4? 13^4=28561; but fourth root ~11.4. If Ω≥4, smallest prime factor p ≤ N^(1/4) < ? Need show p could 11 only. Since p≥11. For N≤17041. 11^4=14641 ≤N, and next prime13: if p≥13 then N≥13^4=28561 impossible. Thus p=11. Nice. Then N divisible by11 iff m≡0 or1 mod11. For m=122..131, only122. Factor Ω3. Therefore no. Even m=122 N factor 11*17*79 exactly three. This is elegant.

At m=132, N=11³*13. Note 132≡0 mod11. Need show exact.

Let's ensure lower bound derivation if "product of at least four primes" potentially means product expression can include 1? No, primes. If number could be negative? m positive, polynomial always positive (discriminant, min around10.75). Fine.

We should reason step-by-step. We can explain Ω perhaps total number prime factors with multiplicity.

Potential issue: "at least four not necessarily distinct primes" could mean N has factorization into ≥4 primes, i.e. Ω(N)≥4. If N includes more, lower bound all prime factors at least11. Correct.

Need show no prime factors 2,3,5,7:
- N odd because m(m-1) even.
- Mod3, m(m-1) is? For m residues 0,1,2, products0,0,2; plus 11≡2 gives2,1,1, none0.
- Mod5 as above; perhaps state checking m mod3,5,7 gives no zero. We could combine table of N mod q:
q=3: residues 2,1,1.
q=5: for m0..4 N=m²-m+1 mod5:1,1,3,2,3.
q=7: N≡ m²-m+4:4,4,6,3,3,6,4.
All nonzero. Or show discriminants? Direct table.

Then all prime divisors ≥11.

Suppose Ω≥4. Let p1≤...≤pk, k≥4. N=∏ p_i≥p1^4≥11^4=14641. This gives N≥14641. Since N(m) increasing for positive m (derivative/discrete N(m+1)-N(m)=2m>0), and N(121)=14531<14641, m≥122. Actually N(122)=14773. Good.

For contradiction m<132 then 122≤m≤131. N≤N(131)=17041. If N has ≥4 prime factors, its least prime divisor p must satisfy p^4≤N≤17041. p≥11. Could p=11; 13^4=28561>17041, so p=11. Thus 11|N.

Now congruence mod11:
N=m²-m+11≡m(m-1) mod11, so divisible iff m≡0 or1 mod11. In interval122..131, only m=122 (since 121≡0? Let's check 121 mod11=0, 122 mod11=1; next 132 mod11=0). Good.

At m=122, factorization:
N=122²-122+11=14773.
14773 /11=1343, /17=79. So N=11·17·79. All prime, exactly 3 factors, not at least4. Therefore no m<132.

At m=132:
N=17424-132+11=17303. 11³=1331; *13=17303. Exactly four primes counting multiplicity. Thus least.

This is very clean. Need be mindful N increasing: difference N(m+1)-N(m)= (m+1)m+11 - (m(m-1)+11)=2m >0.

Let's test no possibility m lower than121 despite N maybe has a prime factor less than11 due factors? We established none. If factorization includes a prime repeated, all >=11. Fine.

Let's consider wording "product of at least four not necessarily distinct primes" perhaps they might not require N itself's complete prime factorization has at least 4; if N composite and can write product e.g. 11*17*79*1? No 1 not prime. Equivalent Ω≥4. Fine.

Let's prepare final response step-by-step and final line exact format number only. User asks final numerical answer on last line `Final Answer: <number>` number only no units. We should make sure last line exactly "Final Answer: 132". They said number only, no units perhaps means after colon number. Good.

Need include enough step reasoning, not too verbose. We can articulate residue table.

Let's make residue check table:

\[
\begin{array}{c|ccccccc}
m \bmod 7&0&1&2&3&4&5&6\\
N\bmod7&4&4&6&3&3&6&4
\end{array}
\]
For 3 and5 could prose. Maybe table all:
modulus q | values of N mod q as m runs residues
3: 2,1,1
5:1,1,3,2,3
7:4,4,6,3,3,6,4.
Since none zero. And N odd. Thus least possible prime divisor 11. Note prime 2 excluded odd.

Then lower.

At candidate interval, use smallest prime divisor argument:
If N=product of at least4 primes and p is least prime divisor, p^4≤N. Since all p≥11 and N≤17041:
- If p≥13, N≥p^4≥13^4=28561 contradiction.
So p=11.

Actually if N has at least four primes, p≤N^(1/4), yes p^4≤N. Since p is prime divisor. Even if N has other factors not prime? Complete factorization. Good.

Check m=122 factor all primes: 79 prime. 1343=17*79. Yes.

Could there be m=121 and N<11^4 but perhaps includes prime 2 etc? no. Done.

Let's perhaps explicitly note N values monotonic. At m=121:
121² -121 +11 = 14641-121+11=14531. At m=122 14773. So any m≤121 has N≤14531<14641. Positive monotonic.

At m range 122-131, N max17041. If m=122 factor. For m123-131 not div11 because residues2..10. Fine.

At m132 congruent0, factor.

This likely final. But token budget huge, likely straightforward. Need ensure no hidden interpretation requiring "least positive integer m" maybe at least four primes not necessarily distinct but possibly "a product of at least four primes" can include prime factors in any representation and N=prime times 1? no.

Let's investigate if N perhaps for m<122 can be product of four primes including primes less 11 even though N itself not divisible? Impossible. We checked excluded 2,3,5,7. Prime divisors could be e.g. 4? no.

Let's verify mod5 and7 calculations carefully:

N=m²-m+11.
mod3 11=2:
m0:2; m1:1-1+2=2 mod3 (not1! Wait I earlier said products +11. Let's recalc! Critical. m=1 N=11 ≡2, yes. I wrote table 2,1,1 incorrectly. m2:4-2+2=4≡1. So residues [2,2,1], none. Earlier product m(m-1) residues 0,0,2 plus 11≡2 ->2,2,1. Good.
mod5 11=1:
m0 1
m1 1
m2 4-2+1=3
m3 9-3+1=7=2
m4 16-4+1=13=3. [1,1,3,2,3], yes.
mod7 11=4:
m0 4
m1 4
m2 4-2+4=6
m3 9-3+4=10=3
m4 16-4+4=16=2? Wait earlier [4,4,6,3,3,6,4] perhaps wrong! Let's recalc direct:
m=4, N=16-4+11=23 ≡2 mod7 (21 remainder2), not3.
m=5:25-5+11=31 ≡3.
m=6:36-6+11=41 ≡6.
m=0:11≡4; m1=11≡4; m2=13≡6; m3=17≡3; m4=23≡2; m5=31≡3; m6=41≡6. No zero. Earlier formula product residues m(m-1): [0,0,2,6,12=5,20=6,30=2], +4 => [4,4,6,10=3,16=2,24=3,34=6]. Correct table [4,4,6,3,2,3,6].
So if table, correct. No issue, but should avoid erroneous residues.

Could prove nondivisibility via discriminants:
For odd prime q, N divisible roots if discriminant (-43) square mod q. For q=3 D=-43≡2 non-square; q=5 D=2? -43≡2 non-square; q=7 D=-1≡6 non-square. This is elegant but direct residue table accessible. Need ensure mod q=2 separately. Use table perhaps.

At q=3: D=-43 ≡2, nonsquare.
q=5: -43 ≡2, nonsquare mod5 (squares0,1,4).
q=7: -43 ≡ -1=6, nonsquare mod7 (squares0,1,2,4).
This establishes no roots for q=3,5,7 because leading coefficient1 and quadratic. Simpler than table, but some readers may accept. We can state checking residues. Let's use explicit congruence sets to be rigorous.

N odd: m²-m=m(m-1) even.

Modulo:
- mod3, m(m-1) ∈{0,2}; N ≡ that+2 = {2,1}.
- mod5 products residues {0,2,1}; +1={1,3,2}.
- mod7 products {0,2,6,5}; +4={4,6,3,2}; none0. This concise.

Then min prime divisor 11. Actually prime divisor could be? Yes primes less11 are 2,3,5,7, all excluded.

Let's consider exact factoring at 132: 11^3*13. 1331*13:
1331*10=13310 +3993=17303. correct.
At 122 17*79:
17*79=1343; *11=14773. correct.
m=121 N=14531. no need factor.

We can formulate contradiction:
Assume m<132 satisfies. From lower N≥14641 and monotonic, m≥122. So m∈[122,131].
For such m, N≤N(131)=? recurrence or direct 131²=17161; -131+11=17041. If p smallest prime divisor, p^4≤N≤17041. p cannot be <11. 13^4=28561, so p=11. More directly p≤floor(17041^(1/4)) =11 (since 11^4≤...<12^4 and primes). Then divisibility condition.
N mod11 = m²-m (because11 term) =m(m-1). For m=122..131, m mod11 =1..10; product zero only at residue0 or1, so only122. Wait m residues: 122=1 through131=10, yes only122. m=121 residue0 outside, m132 residue0.
Then N122 only Ω3. Thus none.

At m=132 works. Great.

Need comply "Please reason step by step". We'll number steps.

Final answer last line. No extra after.

Let's draft in final:

Let Ω(n)...
1. Exclude small primes.
2. Bound m below.
3. narrow and rule out.
4. verify.

Use mathematical clarity.

Potential subtlety: If at least four primes but not necessarily complete? We define Ω. State counting multiplicity.

Let's ensure table residues mod3 includes no zero. We can simply say direct check. Maybe:
\[
N \bmod3\in\{1,2\},\quad N\bmod5\in\{1,2,3\},\quad N\bmod7\in\{2,3,4,6\}.
\]
Let's verify mod7 set {2,3,4,6}; yes. So no2,3,5,7.

One concern: Every prime divisor ≥11, but "at least four primes" could include factors 1? no.

Let's final around 500 words.


Question: Tetrahedron $ABCD$ has $AD=BC=28$ , $AC=BD=44$ , and $AB=CD=52$ . For any point $X$ in space, define $f(X)=AX+BX+CX+DX$ . The least possible value of $f(X)$ can be expressed as $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ .

Since opposite edges are equal, \(ABCD\) is a disphenoid. Place its vertices as alternate corners of a rectangular box centered at \(O\):

\[
A=(p,q,r),\quad B=(p,-q,-r),\quad C=(-p,q,-r),\quad D=(-p,-q,r).
\]

Then

\[
AB^2=4(q^2+r^2)=52^2,\quad AC^2=4(p^2+r^2)=44^2,\quad AD^2=4(p^2+q^2)=28^2.
\]

So

\[
q^2+r^2=676,\quad p^2+r^2=484,\quad p^2+q^2=196.
\]

Adding and dividing by \(2\),

\[
p^2+q^2+r^2=\frac{676+484+196}{2}=678.
\]

Also \(A+B+C+D=0\), so \(O\) is the centroid, and

\[
OA=OB=OC=OD=\sqrt{678}.
\]

Now \(f(X)=AX+BX+CX+DX\) is convex, and at \(O\),

\[
\nabla f(O)=\sum_{V\in\{A,B,C,D\}}\frac{O-V}{OV}
=-\frac{A+B+C+D}{\sqrt{678}}=0.
\]

Thus the minimum occurs at \(O\), giving

\[
f_{\min}=4\sqrt{678}.
\]

Since \(678=2\cdot3\cdot113\) is squarefree, \(m=4\) and \(n=678\), so

\[
m+n=682.
\]

Final Answer: 682

Explanation:
We need solve geometry. Need step by step final. Need reason. Let's think thoroughly.

Problem: Tetrahedron ABCD has opposite edges equal: AD=BC=28, AC=BD=44, AB=CD=52. This is an isosceles tetrahedron (disphenoid). For point X in space minimize sum distances to vertices (geometric median / Fermat point for 4 points in 3D). Need least possible value m√n. Find m+n.

We need solve. For tetrahedron with all pairs opposite edges equal, vertices can be represented as (±a, ±b, ±c) with an even? Let's recall: A disphenoid can be inscribed in rectangular box centered at origin with vertices at (±x,±y,±z) with product of signs maybe +? Opposite edges equal. Specifically four vertices of a rectangular parallelepiped alternate vertices: A=(x,y,z), B=(x,-y,-z), C=(-x,y,-z), D=(-x,-y,z). Then AB length = sqrt((0)^2+(2y)^2+(2z)^2)=2√(y^2+z^2). CD same between C and D = (0,-2y,2z)? C=(-x,y,-z), D=(-x,-y,z): diff (0,-2y,2z), same. AC=2√(x^2+z^2), BD same. AD=2√(x^2+y^2), BC same. Given AB=52 -> y^2+z^2 = 26^2=676. AC=44 -> x^2+z^2=22^2=484. AD=28 -> x^2+y^2=14^2=196. Solve: add first two minus third: (y^2+z^2)+(x^2+z^2)-(x^2+y^2)=2 z^2 =676+484-196=964 -> z^2=482. Then x^2 =484-482=2; y^2=196-2=194. Check y^2+z^2=194+482=676. Good. So x=√2, y=√194, z=√482. Maybe simplify: 482=2*241, 194=2*97. Coordinates all √2 times (1, √97, √241) with signs. Center origin.

Need minimize f(X)=sum distances to these four points. Because of symmetry? The set of four vertices is centrally symmetric? Let's check A=(x,y,z), D=(-x,-y,z)? Not opposite. Actually vertices are four alternate corners of box. Are they invariant under 180° rotations about coordinate axes? Yes. But not centrally symmetric as a set? If point P in set, -P? A=(x,y,z); -A=(-x,-y,-z) not one of listed (listed have sign product +? A +++; B +-- product +; C -+- product +; D --+ product +). -A=--- product - not included. So not centrally symmetric. It is a tetrahedron with symmetry group of rectangle? The center of the box O=(0,0,0) is also the circumcenter and centroid? Average of A,B,C,D = (x+x-x-x, y-y+y-y, z-z-z+z?) Let's compute y: A y, B -y, C y, D -y sum0. z: A z, B -z, C -z, D z sum0. x sum0. Yes centroid origin. Good. Also all vertices equidistant from origin: R^2=x^2+y^2+z^2=2+194+482=678. So O is circumcenter/centroid.

Symmetry group includes 180 rotations about x,y,z axes? Set product sign +. Rotation about x maps (x,y,z)->(x,-y,-z): A->B, B->A, C? (-x,y,-z)->(-x,-y,z)=D, D->C. Yes. similarly. These symmetries imply the geometric median maybe at origin if unique and invariant under group. The objective f is convex (sum of Euclidean norms), invariant under these rotations (since set invariant). The minimizer of a strictly convex function invariant under a group with unique fixed point O is O. Is f strictly convex? Sum of distances to non-collinear points is strictly convex except maybe? Norms are convex; sum strictly convex if points not all collinear, yes I think strictly convex on R^3? Need check: sum of Euclidean distances to affinely spanning points is strictly convex? Each ||X-P_i|| is convex, not strictly globally (linear along rays away from point), but sum over points spanning R^3 is strictly convex because if function linear along segment, each term must be linear along that segment; for a distance term to be linear on segment, the segment must lie on a ray starting at P_i (same direction), impossible for all i if directions to vertices not all same. Since points not collinear, yes strictly convex. Therefore unique minimizer. Since invariant under rotation about x axis with fixed set the x-axis? Wait for a point to be fixed by all three 180 rotations, only origin. Actually rotation about x axis fixes x-axis; about y fixes y-axis; intersection fixed all three is origin. Since unique minimizer must be fixed by each symmetry, it lies on x-axis and y-axis etc -> origin. So X=O is minimizer. Then minimum f=4R=4√678 =? 678= ? 678=2*339=2*3*113=6*113. Squarefree? 113 prime, 6 squarefree, so 678 squarefree? 678 factors 2*3*113, no square. Then m=4,n=678,m+n=682. But wait is it that simple? Need verify: Geometric median for four points at vertices of tetrahedron with no angle ≥120? For Fermat-Weber with equal weights, if at a vertex the angle between unit vectors to other points has magnitude ≤1, then vertex could be minimizer. But by symmetry origin likely. However is sum distances to vertices minimized at centroid/circumcenter? For an equilateral triangle, Fermat point=center and sum=3R? For four points on sphere symmetric maybe. Need be cautious: Invariance under three independent reflections/rotations and strict convexity indeed force unique minimizer at common fixed point if the group action has only origin fixed. Does the objective invariant under 180 rotations about coordinate axes? Need ensure rotations are symmetries of the set of four points. Yes as above. The axes pass through midpoints of opposite edges? Coordinates chosen centered at centroid. Good.

But hold on: The coordinates representation uses A=(x,y,z), B=(x,-y,-z), C=(-x,y,-z), D=(-x,-y,z). Then rotation about z-axis maps (x,y,z)->(-x,-y,z): A->D, D->A, B=(x,-y,-z)->(-x,y,-z)=C, C->B. Good. So all three. The intersection of fixed points of rotations about x,y,z axes is origin only. Strict convexity gives unique global minimizer. Invariance implies if X* unique minimizer, R_x X* also minimizer, hence =X*, so X* on x-axis; similarly on y-axis and z-axis. A point cannot lie on all three axes unless origin. Good.

Then minimal f= sum distances from O to vertices. Each distance sqrt(x^2+y^2+z^2)=sqrt678. Sum 4sqrt678. m+n=682. But wait: 4√678 = √ (16*678)=√10848 = maybe express as m√n with n squarefree. Since 678 squarefree? 678=2*3*113. yes. m=4 n=678. sum 682. Is that too low? Let's verify with possible better point not origin? Strict convex argument seems airtight. But let's test numerically maybe. Coordinates: x=1.414, y=13.928, z=21.954. R=26.038. Sum=104.153. Could there be a point near center with lower? At origin gradient = sum (O-P_i)/|O-P_i| = -(sum P_i)/R=0. Good. Since all distances equal R, gradient zero. Convex => global min. Even simpler: Because centroid at origin and all vertices same distance, origin is stationary; by convexity minimum. No need group. Wait f convex and differentiable at O (O not equal vertex), gradient = sum (O-A)/R = -(A+B+C+D)/R=0. Convex differentiable function with gradient zero at O -> global minimum. Yes. That's even more direct. Need ensure O inside tetrahedron? It is centroid of vertices, yes inside. But differentiability and gradient zero enough for convex function. Sum of Euclidean norms is convex. At O not coinciding with any vertex, differentiable. ∇f(O)=Σ(O-V)/||O-V|| = -(Σ V)/R =0. Hence O is global minimizer. Good.

Need produce step-by-step. But maybe there is an issue: A,B,C,D as assigned: In a tetrahedron with opposite edges equal, there is a rectangular box such that the tetrahedron vertices are four alternate vertices. We can choose coordinates as above. Need derive carefully.

Let's formalize.

Let the common center be O. For a disphenoid, place vertices at
A=(u,v,w), B=(u,-v,-w), C=(-u,v,-w), D=(-u,-v,w). Then compute:
AB^2=(2v)^2+(2w)^2=4(v^2+w^2)=52^2 -> v^2+w^2=676.
AC^2=(2u)^2+(2w)^2=44^2 -> u^2+w^2=484.
AD^2=(2u)^2+(2v)^2=28^2 -> u^2+v^2=196.
Solving gives u^2=2, v^2=194, w^2=482.
Then O=(0,0,0). It is the average of vertices (centroid), and OA=OB=OC=OD=√(u^2+v^2+w^2)=√678.
For f(X)=Σ|X-V_i|, f is convex. At O, ∇f(O)=Σ (O-V_i)/|O-V_i| = -(A+B+C+D)/√678=0. Hence O is a global minimizer. Minimum =4√678. Since 678 squarefree, m=4,n=678, answer=682.

Need maybe prove f convex: sum of norms. Or use triangle inequality lower bound: For any X, AX+BX ≥ AB? Not enough. Could pair opposite vertices: AX+? Hmm maybe there's a direct inequality using symmetry: Since A+B+C+D=0 and all vertices on sphere radius R, for any X, by convexity of norm, (1/4)Σ||X-V_i|| ≥ ||X - (1/4)ΣV_i|| = ||X||? That gives f≥4||X||, not helpful (at X=0 gives 0). Actually Jensen for convex norm: average distances ≥ distance from X to centroid = ||X||; minimization of lower bound not enough. Another inequality: Σ||X-V_i|| ≥? Could use support function. But gradient convexity is fine.

Let's double-check the coordinate assignment matches edge labels: We set AB=52 (given), AC=44, AD=28. Then CD should equal AB: C=(-u,v,-w), D=(-u,-v,w): diff=(0,-2v,2w) length=2√(v^2+w^2)=52 yes. BD: B=(u,-v,-w), D=(-u,-v,w): diff=(-2u,0,2w) length=2√(u^2+w^2)=44 yes. BC: B=(u,-v,-w), C=(-u,v,-w): diff=(-2u,2v,0) length=2√(u^2+v^2)=28 yes. Good.

Solving: v^2+w^2=676; u^2+w^2=484; u^2+v^2=196. Sum all: 2(u^2+v^2+w^2)=1356? 676+484+196=1356; half=678. Good. Then u^2 =678-676=2; v^2=678-484=194; w^2=678-196=482. Equivalent. Good. Earlier solving z^2=482 yes. Sum=678. Good.

Wait check: u^2 = total - (v^2+w^2)=678-676=2. v^2=678-484=194. w^2=678-196=482. Yes.

n squarefree? 678/2=339=3*113. 113 prime (not divisible by 7? 7*16=112; by 11? 11*10=110; sqrt113~10.6, check primes 2,3,5,7; no). So yes.

Need produce final answer with last line exact format 'Final Answer: <number>' number only no units. The user requested step-by-step. We can provide succinct but enough. Need include final numerical answer on last line. Ensure no extra after. Good.

Potential pitfall: The least possible value of f for any point in space: For four points, if one vertex sees the opposite face under angle ≥? There are Weiszfeld conditions. But gradient at centroid zero due to symmetry. Convex ensures global min. Good.

Let's think if f is strictly convex and if O is indeed unique. Not needed for value but if gradient zero for convex function, global minimizer; uniqueness not required. Sum of Euclidean distances to four noncoplanar points is convex; yes. Actually Euclidean norm is convex; finite sum convex. If ∇f(O)=0, then for all X, f(X) ≥ f(O)+∇f(O)·(X-O)=f(O). This uses convexity (subgradient inequality). Good.

Maybe O lies inside tetrahedron? The average of vertices with positive coefficients 1/4, yes inside. But even if not, differentiability at O only needs O not equal a vertex; true. It can be outside; but it's inside.

Let's consider deriving coordinates perhaps with p,q,r to avoid conflict with point X. Use a,b,c as half side lengths of rectangular box. Since problem uses X as variable point, don't use x for coordinate maybe use p,q,r. Let A=(p,q,r), etc. Then p^2=2, q^2=194, r^2=482. Good.

Need mention existence of such coordinates: Because opposite edges are equal, the tetrahedron is a disphenoid; it can be placed as alternate vertices of a rectangular box centered at O. We can state this. If concerned about proving, we can say choose the rectangular box whose faces are perpendicular bisector planes? Hmm. For an answer, it's acceptable. But maybe include a sentence: For a tetrahedron with equal opposite edges, the three segments joining midpoints of opposite edges are mutually perpendicular and bisect each other; equivalently vertices can be written in this form. Maybe overkill. The problem likely from contest; using disphenoid coordinates is fine. But user requested step-by-step, include enough derivation.

Let's see if there is an even more elegant route: In a disphenoid, the centroid=circumcenter. Is that always? Yes. If opposite edges equal, the tetrahedron is isosceles; all faces congruent acute triangles? Let's check side lengths of each face: Faces ABC have edges AB=52, AC=44, BC=28. Face ABD: AB=52, AD=28, BD=44. Same. All faces congruent 28-44-52. Is triangle acute? 52^2=2704, 28^2+44^2=784+1936=2720 >2704, acute barely. Circumcenter=centroid? For disphenoid yes. Then if O is centroid and equidistant from vertices, gradient zero. Could solve circumradius via formula? Coordinates easiest.

Maybe there is formula for circumradius of disphenoid: R^2=(a^2+b^2+c^2)/8? Where a,b,c are the three distinct edge lengths? Let's check: opposite edge lengths are 28,44,52. Coordinates gave R^2=(28^2+44^2+52^2)/8? Sum=784+1936+2704=5424; /8=678. Yes. Then f_min=4R=4√678. Nice. We can present coordinate derivation, maybe mention formula. If use formula, need justify. Coordinates justify.

Let's ensure m,n positive integers and n squarefree: m=4, n=678; m+n=682. Good.

Let's formulate final solution elegantly:

Since AD=BC, AC=BD, AB=CD, ABCD is a disphenoid. Put its vertices at alternate corners of a rectangular box centered at O:
A=(p,q,r), B=(p,-q,-r), C=(-p,q,-r), D=(-p,-q,r).
Then:
AB^2=4(q^2+r^2)=52^2,
AC^2=4(p^2+r^2)=44^2,
AD^2=4(p^2+q^2)=28^2.
Thus q^2+r^2=676, p^2+r^2=484, p^2+q^2=196. Adding and dividing by 2 gives p^2+q^2+r^2=(676+484+196)/2=678.
So O is the centroid because A+B+C+D=0, and OA=...=√678.
The function f is convex. At O,
∇f(O)=Σ_{V=A}^D (O-V)/OV = -(A+B+C+D)/√678=0.
Therefore f attains its global minimum at O. Minimum =4√678.
Since 678=2·3·113 is squarefree, m=4,n=678 and m+n=682.

Last line Final Answer: 682.

Need be careful with gradient expression: (O-V)/|O-V| = -V/R. Sum = -(A+B+C+D)/R=0. Yes.

Let's include "for V in {A,B,C,D}". Good.

No units. Done.

Check desired overbrevity 5: step-by-step enough.

One more subtlety: In the coordinate setup, O=(0,0,0) is center of rectangular box. Is it necessarily the circumcenter? Yes because each vertex coordinate squares sum same. Good.

Let's maybe justify the coordinate placement in one clause rather than prove. Good.

Let's consider if there could be a lower value at a vertex due to nondifferentiability? No because convex and stationary at O. If f is convex and differentiable at O with gradient zero, global min. Even if not strictly, O is a minimizer. Good.

Potential challenge: Sum of distances f is convex but not everywhere differentiable; at O it is differentiable since O is not one of vertices. Good.

Let's formulate final answer. Ensure final line exactly format. No Markdown after. Good. Could use LaTeX. Last line no LaTeX? User says format 'Final Answer: <number>' number only. We'll write `Final Answer: 682`.

Proceed.


Question: Define a sequence recursively by $f_1(x)=|x-1|$ and $f_n(x)=f_{n-1}(|x-n|)$ for integers $n>1$ . Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500,000$ .

Let \(Z_n\) be the set of zeros of \(f_n\), and define
\[
T_n=\frac{n(n-1)}2.
\]

We claim that
\[
Z_n=\{n-T_n,n-T_n+2,n-T_n+4,\ldots,n+T_n\}.
\]

For \(n=1\), this gives \(Z_1=\{1\}\), which is true.

Assume the formula holds for \(f_{n-1}\). Since
\[
f_n(x)=f_{n-1}(|x-n|),
\]
a zero occurs when \(|x-n|\) is a nonnegative zero of \(f_{n-1}\).

The largest zero of \(f_{n-1}\) is
\[
(n-1)+T_{n-1}
=(n-1)+\frac{(n-1)(n-2)}2
=\frac{n(n-1)}2=T_n.
\]
All zeros have the same parity as \(T_n\). Thus their nonnegative values are
\[
\delta,\delta+2,\delta+4,\ldots,T_n,
\]
where \(\delta\in\{0,1\}\) has the parity of \(T_n\). Therefore the zeros of \(f_n\) are
\[
n\pm(\delta,\delta+2,\ldots,T_n),
\]
which form the arithmetic progression from \(n-T_n\) to \(n+T_n\), proving the claim.

Hence the number of zeros is
\[
T_n+1,
\]
and their sum is
\[
n(T_n+1)
=n\left(\frac{n(n-1)}2+1\right)
=\frac{n^3-n^2+2n}{2}.
\]

Now check the threshold:
\[
S_{100}=\frac{100^3-100^2+200}{2}=495100,
\]
while
\[
S_{101}=\frac{101^3-101^2+202}{2}=510151.
\]

Since the sum is increasing, the least such \(n\) is \(101\).

Final Answer: 101

Explanation:
We need solve. Need provide step-by-step. Need deeply understand recursion f1=|x-1|, f_n(x)=f_{n-1}(|x-n|). Need zeros multiset perhaps roots. Need sum. Find least n where sum zeros >500k.

Let's derive.

f_n piecewise absolute nested. Solve f_n(x)=0. Since f_{n-1}(y)=0 for y in zero set S_{n-1}, equation |x-n| ∈ S_{n-1}. If S elements nonnegative perhaps. For each zero r of f_{n-1}, solutions x=n±r. Duplicates if r=0: same. So zero set S_n = {n ± s : s∈S_{n-1}}. Starting S1={1}. Need possibly negative generated if s>n, and then as inputs to abs next. But roots can negative. If duplicates maybe algebraic? Functions absolute compositions zeros distinct. We need sum over zeros likely distinct zeros. Need characterize and sum. S_n is images of binary sign choices applied recursively. Let's expand.

S_2: from s=1 n2 => x=1,3.
S3: for 1,3 -> x=3±1=2,4 and x=3±3=0,6. {0,2,4,6}.
S4: for each -> {4,6}, {2,8}, {0,10}, {-2,12}; distinct all likely. Sum each maps 4±s gives 8, count doubles, sum doubles. Starting recurrence:
A_n=S sum, N=|S_n|. Every root s gives two roots n+s and n-s; same only if s=0 (exists after n≥3!). Ah if s=0, x n twice same; transformation of remaining: each nonzero gives two. Number N_n = 2 N_{n-1} - indicator(0∈S_{n-1}). Sum T_n = 2 n N_{n-1}, including 0 no effect, because pair sums 2n. Is 0 in S_m iff? S3 contains 0; likely 0 persists? Mapping 0 at next gives m, and can generate zero if some prior root equals next m. Does S_{m} include 0? For m≥3 perhaps yes by induction? If S_{m-1} includes m, then. ForS3 has root 6 but not4. S4 no 0? listed {-2,0? Wait S3 roots {0,2,4,6}. At n4: roots 4±0=4, 4±2=2,6, 4±4=0,8, 4±6=-2,10. yes S4 includes 0 from s=4. Indeed all previous S3 includes n=4. Does S_n include n+1 perhaps root |1+...? Max M_n = M_{n-1}+n, starting M1=1 -> n(n+1)/2. Includes n+1 perhaps subset arithmetic constraints. Need inspect parity/range. But likely 0 roots for n≥3? Let's check mapping to n+1 requires s=n+1 in S_{n-1}. Does S_{n-1} include n? For n=4 yes 4. At n5: if S4 contains5 then 0. Does S4: {-2,0,2,4,6,8,10} no5! Wait mapping S3: 4±... did I list 4±6 = -2,10. no 12 because 3-> 0,6 for previous. Thus no 0 at S5 unless another s=5. likely S5 roots n5 ± S4: {3,7,5,1,9,2,8,0,10,-1,11,-3,13}. Let's sorted -3,-1,0,1,2,3,5,7,8,9,10,11,13 (13 roots). It does include0 from s=5? Actually n=5 and s=5 absent. But roots from S4 values: n-s: 7,5,3,1,-1,-3? Let's calculate S4 values [-2,0,2,4,6,8,10] => n+s [3,5,7,9,11,13,15] and n-s [7,5,3,1,-1,-3,-5]. Combined {-5,-3,-1,1,3,5,7,9,11,13,15}, no 0. Ah mapping: S4 includes? Did S4 have 0 yes. x=5 ± 0 =5, duplicate only one. no zero unless s=5, which does exist as output;wait 5 is root, generated by s=0. To get f5 zero requires |x-5| in zero set S4. Is 5 in S4? no. Correct no zero. Could zero periodically. We need count and sums carefully.

Maybe all zero sets satisfy parity same n? Analyze affine signed sequence.

Zeros f_n are values:
Starting r1=1. At stage k=2..n, x? Recursion orientation: root set S_k = {k ± s: s∈S_{k-1}}. Thus each root can be expressed n ± (n-1 ± (... ±1)). Signs each ±, so
x = n + σ_{n-1}(n-1)+...+σ_1 1 perhaps yes, σ_i ∈ ±, with caveat collapsing zero doesn't alter represented assignments. Specifically S_n = { n + ε_{n-1}(n-1)+...+ ε_1*1 : ε_i∈{±1}}. Let's test n4. n=4 plus signed sum 3ε3+2ε2+1ε1, possible totals odd ±? signed 1+2+3=6 even sum always parity 0 because 1+2+3 even, yields -6,-4,-2,0,2,4,6 possible; plus4 => -2,0,2,4,6,8,10 yes. For n5, signed sum 4+3+2+1=10 parity even, values same parity; plus5 odd. Our list all odd. Good. Duplicates subset sums.

So zero set corresponds to all integers of parity n + total T=n(n-1)/2? Signed sum has parity T; x parity n+T. Bounds [-T,T], likely all values in interval with that parity? Important. For weights 1..n-1, can every signed subset sum from -T to T same parity be achieved? Equivalent subset sums 0..T. For weights 1..m, all integers 0..T achievable. Yes. Thus S_n = {n - T, n-T+2, ..., n+T}, where T=n(n-1)/2. Wow check n5 T=10 => [-5,15] odds = list yes. n4 T=6 => [-2,10] even yes. n3 T=3 => [0,6] even yes. n2 T1 => [1,3] even yes. n1 T=0 => {1} special lower=1 and upper1. Formula n±0 yes {1}. Great.

Then zeros are arithmetic progression step 2 from n-T to n+T. Number = T+1 (since length 2T step2), for n≥1. Sum = average n times count = n(T+1) = n[n(n-1)/2+1] = (n^3-n^2+2n)/2. Check S4 count7 sum28? list sum 28, formula 4*(6+1)=28. S5 sum? count11, average5 sum55. list sum? -5-3-1+1+3+5+7+9+11+13+15=55 yes. Easy.

Find least n integer such that (n^3 - n^2 + 2n)/2 >500,000. Need calculate threshold around cube root 1m ~100. For n=100: (1,000,000 -10,000+200)/2=990,200/2=495,100. Less. n=101: 1,030,301 -10,201 +202 =1,020,302 /2=510,151. So n=101. likely answer.

But need ensure sum zeros interpretation perhaps zeros include multiplicity? Distinct same anyway. Need step by step. However be wary: f recursion maybe f_n=f_{n-1}(|x-n|) creates formula f_n(x)= nested abs? We found root interval. Need prove set is contiguous parity. Could use subset sum induction or interval induction. We need present accessible.

Let's inspect possible trap. "sum of zeros" perhaps roots possibly repeated due function pieces? We identified distinct. At n=3 set 0,2,4,6. Sum12 formula n(T+1)=12. Fine.

Could derive signed representation:
f_n(x)=0 iff recursive |...|=1 eventually. Let perhaps signs correspond. Since |a|=b iff a=±b. Unwind:
f_n(x)= f_{n-1}(|x-n|), zero when |x-n| root r ∈ Z_{n-1}; thus x=n±r. Starting root1. This recursively produces n ± (n-1 ± (... ±1)). Every choice signs gives root, but not necessarily unique. Thus set subset of arithmetic interval. Conversely all numbers of right parity in intervalarise because every integer between 0 and T represent subset sum of 1..n-1. Signed sum = 2 subset sum -T. Then x=n+signed. So exact.

Maybe simpler interval induction: At each stage n, Z_{n-1} is interval [lower, upper] step2 symmetric around n-1. Mapping n ± this yields two intervals? n + Z = [n+(n-1)-Tprev etc] and n-Z; because symmetry perhaps their union continuous same parity. Let Tprev=(n-1)(n-2)/2. Center old n-1. n+ old ranges n+(n-1-Tprev) = 2n-1-Tprev to n+n-1+Tprev; n-old ranges n-(n-1+Tprev)=1-Tprev to n-(n-1-Tprev)=1+Tprev. These intervals likely overlap/gap? Lower interval ends 1+Tprev, upper starts 2n-1-Tprev. Difference start-end =2n-2-2Tprev = (n-1)(? 2 - (n-2)) = (n-1)(4-n), not generally overlap for n>4. Yet resulting union magically fills due old set step2? For n5, old S4 [-2..10], n+ old [3..15], n-old [-5..7], overlap [3..7]. fine. Formula I computed Tprev=6, start upper n+(lower old)=5+(-2)=3; other upper n-lower old=7, overlap. General start 2n-1-Tprev; other end 1+Tprev. Compare 1+Tprev - (2n-1-Tprev)=2Tprev-2n+2=(n-1)(n-2)-2(n-1)=(n-1)(n-4). For n≥4 overlap/touch. For n3 start? okay. So interval induction works n≥4. But signed subset proof cleaner.

Wait mapping intervals union and parity perhaps step 2 continuous. Signed representation explicitly all signed sums? Need ensure zero set exactly all sign combinations; although collapse of | at each branch potentially both signs always represent roots, yes.

Let's explicitly derive:
Let Z_n denote set of zeros.
Z_1={1}.
If r∈Z_{n-1}, then f_n(n±r)=f_{n-1}(r)=0. Conversely if f_n(x)=0 then |x-n| is nonnegative zero in Z? Careful roots Z_{n-1} can negative, but |x-n|≥0. Equation |x-n|=r only for nonnegative r. Our recurrence statement Z_n={n±z : z∈Z_{n-1}} appears includes negative z values producing same outputs as positive |z| and still valid because if z negative root, |x-n|=z has no real solutions, so cannot include n±z directly! Ah! Important! Let's revisit. f_n(x)=f_{n-1}(|x-n|). Z_n consists x where |x-n| ∈ Z_{n-1}∩[0,∞), then x=n±r. Negative roots do not directly generate. Earlier mapping all roots n±s is wrong once negative roots appear! Yet formula interval perhaps still maybe mapping relation needs abs: S_n = {n±|s|? Since if |x-n|=s and s nonnegative. For negative s no. Equivalent {n±|s|: s∈S_{n-1}}}, because positive counterpart may not be root but abs equation. Indeed x=n±|s|. Did our sign expansion n±(n-1±(...)) remain valid? Let's test actual sets earlier: We mistakenly applied n± all including negative but for negative s, n+s may still perhaps generated via positive root counterpart? At n5 S4 includes -2. Correct roots from nonnegative S4 {0,2,4,6,8,10}: x=5± those gives {5,3,7,1,9,-1,11,-3,13,-5,15}, same list. Treating all s and n±s gave extra? For s=-2 gives 3,7 already duplicates, so final same coincidentally due ± maps: n±s equals n∓|s|. Ah if include both n+s and n-s for negative, that's exactly n±|s|, so same set. Thus set mapping all signed roots still produces same because pair around n depends only |s|. For any root s, n±s = n±|s| as unordered pair. So valid despite equation no direct x=n+s for sign; the other is valid. Set notation {n±s} means both, and both equal n±|s|. If s<0, both arevalid generated by |s| equation, though s itself need not positive root. Thus okay. But converse: x=n±r for r=|x-n|, which may not be a root if negative? r=|x-n|≥0 and f_{n-1}(r)=0, so r is nonnegative root. Thus yes. Can establish Z_n={n±|z| : z∈Z_{n-1}}. Does signed expansion with arbitrary inner signs hold despite taking absolute at each recursive argument? Let's formulate carefully.

To unwind:
f_n(x)=0 iff |x-n| = a_{n-1}, where a_{n-1} is a nonnegative zero of f_{n-1}. Let x-n= ± a_{n-1}; this ± corresponds to possibly signed cumulative expression.

Then f_{n-1}(a)=0 iff |a-(n-1)| = a_{n-2} nonnegative zero. Thus a-(n-1)= ±a_{n-2}, so a=(n-1)±a_{n-2}. Therefore x=n ± [(n-1) ± a_{n-2}], continuing to a1=1. So yes arbitrary sign combinations. The intermediate a values are nonnegative selected zeros but sign choices produce all? Inductively all signed expressions might include negative values at some intermediate stage, and cannot be arguments. Example expression at stage for f5 root x maybe inner a4 =? f5: |x-5|=a4 nonnegative zero S4. All nonnegative S4 values have representations 4 ± a3 with a3 nonnegative S3. Does that give signs choices yielding all outer sums? S4 includes -2 too irrelevant. The set of nonnegative roots may not be all signed full sums. But our final signed representation x=n + ε_{n-1}(n-1)+...+ε1*1 may include paths where partial suffix (n-1)+... is negative, impossible because at previous stage input absolute must be nonnegative root. Are all such final values nonetheless representable with valid path / sign flips? Need inspect.

Let's derive correctly using absolute mapping and zero sets, not blindly arbitrary. Earlier proposed signed expression all signs might overcount values. Test sets agreed through 5, likely symmetry makes final interval anyway. But proof needs accurate.

At recursion, define R_n perhaps all zero values. Equation:
x = n + ε_n? Let r_{n-1}=|x-n|≥0 and r_{n-1}∈Z_{n-1}. Then x=n±r_{n-1}. Although r may have signed representation n-1 + signs but only if nonnegative. Since zero set Z_{n-1} is symmetric? Our conjectured Z_m centered m and symmetric around m, not around 0. For n-1, Z_{n-1}=[m-Tprev,m+Tprev], can have negatives. Nonnegative portion is [max(0,m-Tprev), upper]. Does every target interval final arise? Let's test recurrence directly from interval, with abs.

Assume Z_{n-1} is full parity interval [a,b]=[(n-1)-Tprev,(n-1)+Tprev]. Nonnegative roots P:
if a<0, {0? parity issue} etc. Mapping n±p. Need prove final full interval [-?]. At n=5 P={0,2,4,6,8,10}, mapping gives lower from p=10 n-p=-5 up etc union full due enough. General yes likely.

Could use induction intervals and show union of n±P fills desired. Need handle parity endpoints, overlap. But signed subset representation can be adjusted: Since Z_{n-1} itself perhaps all signed sums of weights 1..n-2 plus center, and if a representation r negative, -r is also representable by flipping all inner signs, so |r| has representation; x=n±r equals n±|r|. Thus every full sign combination corresponds to a valid root: choose absolute partial recursively. At first outer, for any sign pattern for suffix, r may be negative but replace with |r| and flip sign correspondingly. However recursively suffix may have subpartials invalid, induction? Signed expansion of roots perhaps via absolute values but final x can be represented n plus a signed total T where total is any signed sum. Let's reason.

Recursive root set operation can be expressed:
Z_n = { n + σ |z| : z∈Z_{n-1}, σ=±1}. If by induction Z_{n-1} equals all signed sums centered n-1 (perhaps including any pattern), then n+σ|s| = n + σ(±s), so is n plus ±s. As s ranges symmetric set Z_{n-1}, perhaps resulting set equals n ± Z_{n-1}, i.e. all full sign patterns. Indeed if Z symmetric under negation? Is it? Z_m as interval center m generally not symmetric around0. But mapping uses ±s directly, not |; {n±s: s∈Z_{n-1}} equals {n±|s|} only if for every negative s, +s output corresponds |s| via choosing opposite: unordered pair n±s = n±|s| as a pair, yes no need |s| in Z. For positive same. So operation on set yields same as n±s for all s, even though each branch not individually condition. Formally union over s {n+s,n-s}; if s<0, these are n-|s|, n+|s|, both are valid because r=|s| (but r may not be zero). Wait validity of x=n-|s| requires |x-n|=|s| and f_{n-1}(|s|)=0. We only know s∈Z, not |s|∈Z. If s<0, is |s| necessarily a root? Our sets perhaps yes? S4 negative root -2, +2 root yes. Is zero set symmetric about 0 partially? Maybe all negative counterparts positive due symmetry of signed sums (flip signs), yes if signed representation valid. Circular.

Let's calculate actual recurrence to verify formula at later n where nonnegative issue matters. Z5 [-5,15] odd, negative -5,-3,-1; their positives 5,3,1 all roots yes. Z6 mapping nonnegative roots in Z5 {1,3,5,7,9,11,13,15}; x=6± -> {-9,-7,-5,-3,-1,1,3,5,7,9,11,13,15,17,19,21}; formula T=15, [6-15=-9,21] step2 yes. Negative counterparts all present. Good.

Can prove signed-sum set by recurrence if show Z_{n-1} closed under s→-s (at least negatives) and then operation is n±s. Is Z_m symmetric under negation? Formula interval center m not generally: e.g Z5 [-5,15], positive beyond 15 no. Not globally. But all negative elements have positive counterpart due lower bound perhaps. T_m=m(m-1)/2. Negative values range m-T to -? Their abs up to T-m=m(m-3)/2. Positive interval extends up to m+T >. parity matches. So yes for m≥? all negatives abs in set. Because full formula.

We can instead prove formula directly via induction with interval details.

Let's establish candidate Z_n = n + E_{n-1}, where E_m = { -T_m, -T_m+2,...,T_m}, T_m=m(m+1? Wait weights 1 through n-1 total T=n(n-1)/2). This candidate all sign sums. Does recurrence preserve? Let's algebra.

Assume Z_{n-1}= {n-1 + signed sum weights1..n-2}. Need show Z_n candidate. Actual:
Z_n={n±r: r∈Z_{n-1}, r≥0}.
Want all n+s where s∈[-T_n,T_n] parity.
Decompose s = ±(n-1)+t perhaps signs weights. Candidate full signed sums. Need find r≥0 root such that n±r=n+s, i.e r=|s|. Need show |s| ∈ Z_{n-1} for every s in candidate signed-sum range. Is that true? Example n=5 s=-4? Candidate parity T=10 even, s values even; s=-4, |s|=4, Z4 [-2,10] yes. n=10 s maybe -30, |30| in Z9 [9-36=-27,45] yes; likely. General |s|≤T_n=n(n-1)/2. Need show |s| lies in Z_{n-1} interval [n-1-T_{n-1}, n-1+T_{n-1}], where T_{n-1}=(n-1)(n-2)/2 =T_n-(n-1). Upper U=n-1+Tprev=T_n. Lower L=n-1-Tprev =? n-1 - [(n-1)(n-2)/2] = (n-1)(4-n)/2. For n≥5, L≤0. Any |s|≤T_n=U and ≥0, but if L positive for small n. For n≥4 L≤0, and parity of |s|? s parity T_n. Z_{n-1} values parity (n-1)+Tprev = n-1+T_n-(n-1)=T_n. So yes |s| is in full arithmetic interval if ≥max(0,L). For n≥4 L≤0. Thus induction works beautifully. For n=2,3 check manually. This proves candidate and recurrence.

Alternatively once formula established at n-1, map nonnegative interval. For n≥4 lower≤0; P includes all same parity 0..T_n? Wait nonnegative Z_{n-1} interval from max endpoint appropriate to U. If L≤0, P = {0,2,4,...,T_n} or starting parity maybe T_n parity. Is T_n parity same as 0 only if T_n even; if T_n odd, values in [0,U] with parity T_n are odd, no 0. P={parity p, p+2,...U}. Mapping n±p yields perhaps intervals union fill candidate. Need check. E.g n=5 T=10 even P evens 0..10 -> n± produces odds -5..15 yes. n=6 T=15 odd Z5 odds includes 1,3,...15 (and negative odd), P odds1..15 -> 6± gives evens -9..21. yes. General. Show:
n+P = [n+p, n+T_n] step2
n-P = [n-T_n,n-p].
These two overlap/touch because n-p ≤ n+p, obvious p≥0; combined arithmetic sequence from n-T_n to n+T_n with step? Each has step2; union [n-T,n-p] and [n+p,n+T], leaving gap between n-p and n+p if p>1 (difference 2p). Wait each is step2, endpoints perhaps parity. Example p=1, gap from n-1 to n+1, missing n? Candidate parity excludes n because step2 parity? n candidate endpoints n-T; T odd => parity n+1, so n not candidate. There are values n-1, then n+1 with difference2, no gap in arithmetic progression (difference2). For p=0, n duplicate and contiguous. General p is 0 or1, so difference 2p ≤2, union fills step2 sequence. Yes.

This is robust.

But perhaps much simpler characterize zeros via triangular intervals using known lemma all signed sums. Need avoid invalid signs.

Let's seek elegant direct formulation maybe f_n zeros are integers x satisfying parity/range. We can prove by induction:
Claim for n≥? Z_n = { n+k : k=-T_n,-T_n+2,...,T_n}, T_n=n(n-1)/2. Verify n=1. For induction, use recurrence:
f_n(x)=0 iff |x-n|∈Z_{n-1}.
By induction Z_{n-1} interval I = [n-1-T_{n-1}, n-1+T_{n-1}] step2. Its nonnegative part for n≥4 is [δ,T_n] where δ=0 or1 depending parity T_n, because upper U=T_n. Lower L=(n-1)(4-n)/2≤0. Then x=n±r, r in that set. As r runs δ,δ+2,...,T_n, n+r and n-r together form exactly [n-T_n,n+T_n] with step2 (because overlap at center/difference ≤2). For n=2,3 manually or induction maybe lower nonnegative constraints still works:
n=2: Tprev=0, Z1={1}, upper T2=1, P {1}, n± gives1,3 candidate [1,3]. δ=1.
n=3: Tprev=1, Z2 {1,3}, nonnegative same, upper T3=3, P {1,3}; n± gives2,4,0,6 candidate [0,6] step2. yes. General formula P perhaps [δ,T_n] but n2 lower L=1 =δ; n3 L=0 but actual Z2 min1, δ=T3 parity=1, so [1,3]. Fine. n1 induction not needed. Thus formula all n.

Let's calculate:
T_n=n(n-1)/2.
For Z_{n-1}, arithmetic progression:
lower L=(n-1)-T_{n-1}=(n-1)(4-n)/2.
upper U=(n-1)+T_{n-1}=T_n (nice).
Parity p = (n-1)+T_{n-1} = T_n mod2. Let δ=T_n mod 2 (0/1). The nonnegative members of Z_{n-1} are r=δ,δ+2,...,T_n, except perhaps lower L >δ? For n=2 L=1 δ=1; n=3 L=0 δ=1, actual min1 =δ. For n≥4 L≤0 so δ. So yes.

Then roots at x=n±r. n-r runs n-T_n,...,n-δ; n+r runs n+δ,...,n+T_n. Since δ≤1, these combine into a step-2 progression over full range. Note parity: both endpoints same parity; all increment2. If δ=1, first ends n-1 and second starts n+1, difference2, no omission relative step2. Good.

Count: (2T_n)/2+1=T_n+1. Sum arithmetic progression = count * average endpoints /2 = (T_n+1)*[(n-T)+(n+T)]/2 = n(T+1).

Then threshold.

Check arithmetic:
S_n=n(T_n+1)=n(n(n-1)/2 +1)=(n^3-n²+2n)/2.
n=100: n²=10k, cube1m. numerator 1,000,000 -10,000 +200=990,200; /2=495,100.
n=101: cube? 101^3=1,030,301; square10,201; 2n202. subtract+ =1,020,302; /2=510,151.
Need show least: S_n strictly increasing positive n; difference S_n-S_{n-1}? Compute maybe:
T_n+1 etc. derivative positive; integer. S_n - S_{n-1}:
[n(n-1)/2+1]n - [(n-1)(n-2)/2+1](n-1).
Could state cubic increasing for n≥1. Verify maybe values. Or threshold directly check n100 below, n101 above and cubic increasing.

Let's not overlook possibility zeros could be non-integers? Absolute nested functions zero conditions lead integer roots as shown. Fine.

Let's inspect recursion notation: f_n(x)=f_{n-1}(|x-n|). For f1 zero x=1. We correctly use f_{n-1}'s zero at nonnegative input. Formula count grows triangular ~n²/2, sum~n³/2. Threshold 100. Seems perhaps contest question designed.

But let's test explicit f2 etc:
f2=||x-2|-1|. zeros |x-2|=1 =>1,3.
f3= f2(|x-3|)= || |x-3|-2|-1|. zeros when ||x-3|-2|=1 => |x-3|=1 or3 =>2,4,0,6. formula.
f4 zeros |x-4| in {0,2,4,6}: 4, 2,6,0,8,-2,10. formula. Great.

Potential ambiguity: "sum of the zeros" could count zeros with multiplicity. All roots distinct set and function piecewise linear; at cusp roots multiplicity one? At x=some perhaps left/right slopes? Multiplicity likely no. No issue.

Need provide step-by-step final numerical answer exact requested last line format number only. We can give equations and explanation. Last line exactly "Final Answer: 101". Ensure no content after.

Let's develop polished proof.

Let Z_n zero set, define T_n. Claim.

Base n=1: T1=0, interval [1,1].

Inductive detail:
Suppose
Z_{n-1}=\{n-1+k: k=-T_{n-1},-T_{n-1}+2,\ldots,T_{n-1}\}.
Then upper endpoint U=n-1+T_{n-1}=n-1+(n-1)(n-2)/2 = n(n-1)/2=T_n.
Lower endpoint L=n-1-T_{n-1}=(n-1)(4-n)/2. It may be positive only for n=2; for n=3 L=0 but parity means actual progression from0 step2: {0? Wait Z2 formula [1,3], not [0,3]. Let's recompute n=3 induction using n-1=2: T2=1, lower=2-1=1, not formula (n-1)(4-n)/2 =2*(1)/2=1. I mistakenly said L0. For n=3 L=1. Candidate T3=3 odd δ=1, nonnegative roots start1. Good. n=2 n-1=1 T1=0 lower1. δ=T2=1. Fine. n≥4 L≤0. So nonnegative roots all δ,δ+2,...T_n? For n=3 yes {1,3}; n2 {1}. For n4, Z3 [0,6], T4=6 even δ0 => {0,2,4,6}. Great.

But what if lower endpoint positive for n=2 and equals δ. n=3 lower1 equals δ. General lower for n=2,3 =1; δ=1. Fine.

Need ensure lower endpoint L itself maybe nonnegative and not equal δ for larger? n≥4 ≤0. So P exact.

Parity p = T_n mod2, since all Z_{n-1} entries congruent n-1+T_{n-1}=T_n mod2. Thus nonnegative members are δ, δ+2,...,T_n. Why if L negative, yes starts δ. If L maybe not integer? It is integer endpoint. Good.

Then equation f_n(x)=0 iff |x-n|=r for r in P (because argument absolute nonnegative). Hence x=n-r or n+r. Sets:
A={n-T_n,n-T_n+2,...,n-δ}
B={n+δ,n+δ+2,...,n+T_n}.
If δ=0, union is [n-T,n] and [n,n+T], overlapping center. If δ=1, first ends n-1, second starts n+1, so consecutive in step-2 progression. Thus Z_n desired. (Could A ordering if δ etc yes.)

This is clean and handles absolute.

Then count/sum.

Let's calculate monotonic proof perhaps state \(S_n=n(T_n+1)\) clearly increases for n≥1 because both factors positive and increasing (T increases). Actually T_n strictly increasing for n≥2, n too, so product increasing. Good.

At n100 and101.

Let's double-check sum formula if zeros progression symmetric around n, yes.

Potential issue n=1 induction count T1+1=1 sum1.

Threshold:
100 sum 495,100, less by4,900.
101 sum510,151. Thus least.

Need "reason step by step" likely deliver.

Let's consider alternate issue: Could "zeros" mean all x where f_n=0 but f_n has domains? all real. Fine.

Let's maybe mention distinct zeros. Our set construction no multiplicity.

Let's ensure no hidden issue due nested absolute function and zeros at boundary maybe duplicate; set handles.

Let's explore if sum might exceed500,000 earlier due roots signed cancellation? We have positive average n and all interval symmetric. n=100 sum495100. correct.

Let's see if recurrence index perhaps f_n(x)=f_{n-1}(|x-n|) with f_{n-1} zeros formula center n-1, yes. We used. Some might derive f_n(x)= nested |...| perhaps zeros interval. Good.

Let's validate random n=6 actual recurrence from Z5 includes negative -5,-3,-1 but ignore; P1,3,...15; roots6± -> 5,7,3,9,1,11,-1,13,-3,15,-5,17,-7,19,-9,21 sorted [-9,-7,-5,-3,-1,1,...21], formula T15 count16 sum96. Formula n(T+1)=6*16=96. List symmetry. Good.

Now answer concise but enough.

Need final numerical answer last line exact. No units. We can have final line.

Let's formulate:

1. Let \(T_n=...\).
2. Lemma.
3. Induction.
4. Sum.
5. threshold.

Use \(\delta\in\{0,1\}\).

Let's write base/inductive assumption accurately.

At induction step, prior set:
\[
Z_{n-1}=\{n-1-T_{n-1}, n-1-T_{n-1}+2,\ldots,n-1+T_{n-1}\}.
\]
Its largest element:
\[
n-1+T_{n-1}=n-1+\frac{(n-1)(n-2)}2=\frac{n(n-1)}2=T_n.
\]
All elements have parity of \(T_n\). Let δ parity. Nonnegative elements? Need be precise for n maybe lower endpoint >δ. We can simply check n=2,3 separately, and for n≥4 lower≤0. But induction base perhaps n1, step2-3 manual. We can state direct checks and for n≥4. Or prove claim via subset sums more elegantly perhaps avoids case details. But interval proof with note.

Let's inspect nonnegative part if lower endpoint is e.g n=3 lower1, δ=T3=1. n2 lower1 δ=T2=1. So formula actually holds all n≥2. We can assert because lower endpoint equals1 for n=2,3 and is≤0 for n≥4. Let's verify lower n4:3-3=0. yes.

Then root sets.

Potential concern when δ=1 and intervals A and B: A n-T through n-1 step2; B n+1 through n+T. Full progression endpoint parity n-T; e.g n-T maybe. Difference from n-1 to n+1=2. good.

At n2 δ1: A from1 to1, B3 to3. full [1,3]. n3 T3=3: A 0 to2? n-T0, n-δ2, step2 =>0,2; B4,6. yes.

Count progression terms:
((n+T)-(n-T))/2+1=T+1. T integer.
Sum:
number*(first+last)/2=(T+1)n.

Threshold.

No need include all explicit f examples, but perhaps helpful. Desired oververbosity5. Step-by-step enough.

Let's consider whether sum of zeros might include duplicate n+δ etc. We count set. If functions have root multiplicity perhaps δ? At each cusp, no duplicates in distinct zero set. The question likely means values.

Let's now think whether there is any chance formula should use n? Let's independently derive explicit f_n perhaps sum roots might be something else if "zeros" includes zeros of f_n as function roots and negative/positive. We did.

Let's inspect f_n expression via composition:
f3(x)=f2(|x-3|)=||x-3|-2|-1|? f2(y)=||y-2|-1| = | |y-2| -1|? Actually f2(y)=f1(|y-2|)=||y-2|-1|. yes.
Zeros f3: | |x-3|-2|=1 => |x-3|=1,3. yes.
General zeros as nested equations.

Everything fine.

Let's ensure threshold calculation:
For n=100 T=4950, count4951, sum=495100. Ah formula n(T+1)=100*4951=495100.
For n101 T=5050, count5051, sum=101*5051 = 505100+5051=510151. Nice perhaps easier. Exceeds. Show increasing due T.

Let's prepare final answer with equations.

Need last line exact `Final Answer: 101` (number only after colon). They specify format '(number only, no units)' likely means number field only, this exact.

Let's draft:

Let Z_n denote distinct zeros. Claim triangular progression.

Maybe include induction proof with δ.

At base n=1. Suppose. Define δ. The nonnegative roots of f_{n-1} are...
Explain largest root equals T_n. Lower endpoint. For n=2,3 lower equals1; n≥4 lower≤0, so set starts δ. All roots parity T_n. Then recurrence.

Actually if lower endpoint for n=4 equals0 and δ=T4=0. For n5 lower=-2 and δ=T5=0. Good.

Count etc.

Let's avoid saying lower endpoint equals1 for n=2,3 if induction starting perhaps T0 no. Fine.

We can state:
The lower endpoint is \((n-1)(4-n)/2\), which is 1 for n=2,3 and ≤0 for n≥4. Since all entries have parity T_n, nonnegative entries are exactly δ,δ+2,...,T_n. Check n3 lower1 δ1. Correct. n2 prior Z1 formula lower=1 δ1.

At induction assumption n-1, for n=2 uses base.

Then roots x=n±r.

No issue.

Monotonic: both n and T_n+1 increase.

Let's now produce final.


Question: Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.

Let \(N\) be the number of people at the concert before the bus arrived.

1. Since adults made up \(\frac{5}{12}\) of the crowd, \(N\) must be divisible by \(12\).
2. After \(50\) people arrived, adults made up \(\frac{11}{25}\) of \(N+50\), so \(N+50\) must be divisible by \(25\).
3. Because \(50\) is divisible by \(25\), \(N+50\) is divisible by \(25\) exactly when \(N\) is divisible by \(25\).
4. Thus \(N\) must be divisible by both \(12\) and \(25\), so the smallest positive possibility is
   \[
   N=\operatorname{lcm}(12,25)=300.
   \]
5. The final crowd then has \(300+50=350\) people, of whom
   \[
   \frac{11}{25}\cdot 350=154
   \]
   are adults.

This is achievable: initially there were \(125\) adults, so the bus carried \(29\) adults and \(21\) non-adults.

Final Answer: 154

Explanation:
We need solve. Need carefully interpret bus 50 more people composition unknown nonnegative integers, initial crowd adults 5/12, after total +50 adults fraction 11/25. Need minimum number adults after. Variables initial total N multiple 12, initial adults 5N/12. Bus 50, b adults integer 0..50. Final total N+50 multiple25 and final adults=11/25(N+50)=5N/12+b. Find minimize A_final. N perhaps positive, all adults? Could zero adults? "Adults made up..." yes integers. Bus carrying 50 more people, perhaps number adults bus. Need solve congruences.

Let initial N=12k, initial adults=5k. final total=12k+50=25m perhaps. final adults=11m. Difference bus adults 11m-5k between 0 and50. But if total final 25m and difference totals automatically b_total50. Congruence 12k+50 divisible25 => 12k+0? 50 ≡0 mod25, so 12k divisible25; gcd 12,25 => k divisible25. k=25t. Initial N=300t, adults=125t. final total=300t+50, final adults=(11/25)(300t+50)=132t+22. Bus adults=7t+22. Constraint <=50: 7t+22 <=50 -> t<=4; >=0 t maybe nonnegative. Min final adults t? t positive crowd perhaps t=0 gives initial zero people, concert crowd? Could k=0 means no people before, problematic. Usually number positive, t≥1, min 154. Is there other modular? Adults after monotonically t. likely 154. But check perhaps bus carrying 50 more people could include children. t=4 max but minimum t=1. Initial total 300, adults125, final total350, final adults154, bus adults29. 29/50 yes. Need reason step by step. Potential alternate setup x initial total. We should articulate divisibility and integrality constraints.

Question minimum after bus. Let final adults A perhaps derive:
Original total n. n divisible 12. n+50 divisible25. Mod 25 n+50 ≡ n (since 50 divisible25), so n divisible25. Since n divisible 12 and 25, n divisible 300. Let n=300kk≥1. Final adults 11/25(300k+50)=132k+22. Min k=1 =154. Need verify bus adult count integer: final 154 -125=29, valid. Ah no need constrain? For k greater bus adult =7k+22 <=50, so k ≤4, but minimum remains1. We can show consistency. Note if k arbitrary, bus has 7k+22 adults among 50, so physically k<=4. k=1 works. Thus minimum.

But possible initial crowd total divisible both fractions. At initial, adult count requires 12 divides N. At final adult count requires 25 divides N+50. As above N multiple25. gcd 12,25, N lcm300. Positive minimum N=300. Adults initially125. Final total350; adult 11/25*350=154. Bus carried29 adults and21 children, valid. Answer exact format last line. They request reason step by step. Keep concise but enough. Ensure no units and last line exact.