Question: What is the sum of the solutions to the equation $\sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}$ ?

Let \(y=\sqrt[4]{x}\), so \(y\ge 0\) and \(x=y^4\). The equation becomes

\[
y=\frac{12}{7-y}.
\]

Multiplying by \(7-y\),

\[
y(7-y)=12,
\]

so

\[
y^2-7y+12=0.
\]

Factoring gives

\[
(y-3)(y-4)=0,
\]

hence \(y=3\) or \(y=4\). Both satisfy the original equation.

Thus the corresponding values of \(x\) are

\[
x=3^4=81 \quad\text{and}\quad x=4^4=256.
\]

Their sum is

\[
81+256=337.
\]

Final Answer: 337

Explanation: We need solve. Need reason step by step and final exact. Let y= fourth root x, likely real principal y>=0. Equation y=12/(7-y), denominator nonzero. Multiply y(7-y)=12 => 7y-y^2=12 => y^2-7y+12=0 => y=3 or4. Both nonnegative. x=y^4:81,256 sum337. Need explain perhaps no extraneous. Final last line exact. Ensure format number only after colon.


Question: The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence.

Each term is a sum of distinct powers of \(3\), so its base-\(3\) representation uses only the digits \(0\) and \(1\).

There is a correspondence between the \(n\)-th binary number and the \(n\)-th term:

- Write \(n\) in binary.
- Interpret those same \(0\)-\(1\) digits as a base-\(3\) number.

For example,
\[
6=110_2 \longmapsto 110_3=12,
\]
which is indeed the sixth term.

Now,
\[
100=1100100_2.
\]
Therefore, the \(100^{\text{th}}\) term is
\[
1100100_3=3^6+3^5+3^2=729+243+9=981.
\]

Final Answer: 981

Explanation:
We need answer step by step. Need solve. Sequence sums distinct powers 3 = numbers with ternary digits only 0/1. Increasing. 100th corresponds to binary representation of 100 interpreted in base 3. 100 decimal binary 1100100 (64+32+4), then interpret digits in base3: 3^6+3^5+3^2=729+243+9=981. Need carefully indexing: 1st = 1 corresponds binary 1. nth term write n binary then read base3. n=100 => 981. Need explain perhaps derive via blocks/ranges. User requests reason step by step and final numerical last exact. Ensure last line format. Let's formulate moderately. Maybe mention no zero. Powers 3^0 etc. Distinct sums. Mapping.

We should be aware sequence given 1,3,4,9,10,12,13. Indeed binary:
1 ->1
2 ->3
3 -> 11 base3=4
4 ->100=9
5=10
6=12
7=13.
100 decimal binary =1100100. Base3 1*729 +1*243+0*81+0*27+1*9+0*3+0 =981.

Need perhaps establish one-to-one and increasing. Binary digit subset map is order-preserving between integer binary and ternary because same positions and base 3 > 2. Or use first 2^k terms. Could count 100. For completeness step-by-step.

Maybe derive by block:
First 1 term uses exponents <1? 1.
For k? Numbers involving powers 3^0 through 3^{k-1}: 2^k-1 terms. Up to 3^5? 2^6-1=63. Thus 64th is 3^6=729. Then 100-64=36th term among numbers with leading 3^6. Need append ternary 0/1 lower bits represented by binary 36=100100 => positions 5,2 ->243+9 =>981. This is intuitive. But direct map simplest.

Let's make step-by-step:
1 each selected subset of powers corresponds coefficient sequence 0/1 in base 3.
2 Map n's binary digits to same coefficient sequence base3; 1st n=1.
3 100 in binary 1100100_2.
4 interpreted as base3: 1100100_3 = 729+243+9=981.
Need possibly show it is indeed increasing mapping. Any larger n has higher or same binary length / lexicographic digits, and interpreting in base 3 preserves order. State.

Last line exactly Final Answer: 981. Ensure no text after.


Question: Let $S$ be the sum of the base $10$ logarithms of all the proper divisors of $1000000$ . What is the integer nearest to $S$ ?

We have
\[
1{,}000{,}000=10^6=2^6\cdot 5^6.
\]
Thus its number of positive divisors is
\[
(6+1)(6+1)=49.
\]

The product of all positive divisors of a number \(n\) with \(d(n)\) divisors is
\[
n^{d(n)/2}.
\]
Therefore, the product of all divisors of \(1{,}000{,}000\) is
\[
(10^6)^{49/2}.
\]

Since “proper divisors” excludes \(1{,}000{,}000\) itself, their product is
\[
\frac{(10^6)^{49/2}}{10^6}=(10^6)^{47/2}=10^{141}.
\]

Hence the sum of their base-\(10\) logarithms is
\[
S=\log_{10}(10^{141})=141.
\]

Final Answer: 141

Explanation:
We need solve. Need reason step by step final exact nearest integer. Need parse proper divisors ambiguity: proper divisors often all positive divisors excluding n (may include 1), perhaps sum logs. n=1,000,000=10^6=2^6 5^6. Number divisors 49. Proper excluding n =48. Sum logs all proper divisors. Product divisors = n^{d(n)/2}=n^{24.5}; excluding n product proper = n^{23.5}. S=23.5 log10 n =23.5*6=141. Nearest 141 exactly. Is there possibility "proper divisors" excludes 1 too? Then divide by 1 no impact! So same sum logs because log1=0. Thus 141. Straight. Need step-by-step perhaps derive symmetry. Check logs can negative none. Divisors powers 2^a5^b, sum exponents. Formula.

Could mention all positive divisors product. Number 49. Pairing d and n/d; odd square and n perfect square means central sqrt(n), product total n^{49/2}. Removing n gives n^{47/2}? Wait careful: product all divisors formula n^{d/2}=n^{24.5}; product proper excluding n = n^{24.5}/n = n^{23.5}; exponent 47/2 yes. log n=6 => 141.

Nearest integer itself 141. Final last format. Need comply numerical only last. Could provide detailed.


Question: The shortest distances between an interior diagonal of a rectangular parallelepiped , $P$ , and the edges it does not meet are $2\sqrt{5}$ , $\frac{30}{\sqrt{13}}$ , and $\frac{15}{\sqrt{10}}$ . Determine the volume of $P$ .

Let the side lengths of the rectangular parallelepiped be \(a,b,c\), with vertices
\[
(0,0,0)\quad\text{and}\quad(a,b,c)
\]
as the endpoints of an interior diagonal. Parametrize the diagonal by
\[
D(t)=(at,bt,ct),\qquad 0\le t\le 1.
\]

Consider an edge parallel to the \(x\)-axis that the diagonal does not meet:
\[
E(s)=(s,b,0),\qquad 0\le s\le a.
\]
The squared distance between \(D(t)\) and \(E(s)\) is
\[
(s-at)^2+b^2(1-t)^2+c^2t^2.
\]
At the minimum, \(s=at\), and minimizing the remaining expression gives
\[
t=\frac{b^2}{b^2+c^2}.
\]
Thus the shortest distance to an edge parallel to \(a\) is
\[
d_a=\frac{bc}{\sqrt{b^2+c^2}}.
\]
Similarly,
\[
d_b=\frac{ac}{\sqrt{a^2+c^2}},\qquad
d_c=\frac{ab}{\sqrt{a^2+b^2}}.
\]

Taking reciprocal squares gives
\[
\frac1{d_a^2}=\frac1{b^2}+\frac1{c^2},
\]
and analogous equations for the other two directions. Hence, after relabeling if necessary,
\[
\frac1{b^2}+\frac1{c^2}=\frac1{20},
\]
\[
\frac1{a^2}+\frac1{c^2}=\frac{13}{900},
\]
\[
\frac1{a^2}+\frac1{b^2}=\frac{2}{45}.
\]

Writing these fractions with denominator \(900\), we obtain
\[
\frac1{a^2}=\frac{40+13-45}{2\cdot900}=\frac1{225},
\]
\[
\frac1{b^2}=\frac{40+45-13}{2\cdot900}=\frac1{25},
\]
\[
\frac1{c^2}=\frac{13+45-40}{2\cdot900}=\frac1{100}.
\]
Therefore,
\[
a=15,\qquad b=5,\qquad c=10,
\]
so the volume is
\[
abc=15\cdot5\cdot10=750.
\]

Final Answer: 750

Explanation:
We need solve. Need reason stepwise, final exact number likely. Need understand geometry shortest distances between an interior diagonal (space diagonal?) of rectangular parallelepiped and edges it does not meet. A rectangular box dimensions a,b,c. Choose a space diagonal, connects opposite vertices. It does not meet which edges? There are 12 edges; diagonal endpoints incident to 6 edges total, leaving 6 edges it doesn't meet? But perhaps distances come in pairs: For a fixed diagonal, six non-incident edges; distances likely grouped into 3 pairs, corresponding to opposite pairs, with three distinct distances. Given three. Need derive distance between a diagonal line segment and edge, likely formula based dimensions, then solve.

Let coordinate box [0,a]x[0,b]x[0,c], diagonal from O=(0,0,0) to D=(a,b,c). Edges incident O or D meet segment. Remaining edges: x-direction edges at (y,z) possibilities (b,0) and (0,c)? Let's enumerate nonincident.
Edge directions:
x at y,z in {0,b}×{0,c}: incident O if (0,0); D if (b,c). Other (b,0),(0,c).
y: (x,z): nonincident (a,0),(0,c).
z: (x,y): (a,0?) incident? z edge at (a,b) incident D, at 0,0 O; remaining (a,0),(0,b).
For each direction, perhaps pair symmetric and distances same? x edges b,0 vs 0,c under complement x->a-x etc yes likely same. Distances d_a? An edge parallel x, line through (0,y0,z0), versus space diagonal param t(a,b,c). Need closest points within segments (parameters x/a and y/b etc). Distance between skew supporting lines may have closest points interior, likely depends ratios. Formula.

Given distances values:
2√5 ≈4.4721
30/√13≈8.3205
15/√10≈4.7434.
Could map to dimensions pair perhaps. Solve equations.

For infinite line diagonal direction v=(a,b,c) and x-edge u=(1,0,0) through P=(0,b,0) (or 0,0,c). Distance between skew lines:
| (P-O) · (u × v)| / |u × v|.
For u=x axis: u×v=(0,-c,b). For P=(0,b,0): dot = -bc? absolute bc. Norm √(b²+c²). Thus infinite line distance = bc/√(b²+c²). Nice. For each edge parallel x. Similarly y: ac/√(a²+c²); z: ab/√(a²+b²). But these are distances between supporting lines, and foot perhaps inside both segments? Need check. The closest line points parameters. For x edge, from O diagonal point t(a,b,c), edge point (s,b,0). Solve orthogonality to v and u:
difference edge - diag perhaps w. Equations s-at =0 for x; v dot (P+s x - tv)=0. s=at, then at*? v dot P = ab; v dot (s,0,0)=a s=a² t, so t(v²)= ab +? Wait P=(0,b,0), diag tv. Difference P+s u - tv; orthog u gives s-at=0. Orthog v: ab + as - tL²=0 -> ab+a²t-tL²=0 => t=ab/(L²-a²)=ab/(b²+c²). For inside t ∈ [0,1] requires ab ≤ b²+c². Not always. If no, closest occurs endpoints perhaps. But likely condition maybe dimensions arranged. For other x edge P=(0,0,c), t=ac/(b²+c²). Symmetric pair maybe one? Both line distance same but closest t may only one or both within depending ab, ac. Segment distance could differ! Yet symmetry complement diagonal? Let's investigate. Pair x edges (0,b,0) and (0,0,c): diagonal has endpoints O,D. Distances perhaps formulas include whether t <=1:
for edge (b,0): t=ab/(b²+c²)
for (0,c): t=ac/(b²+c²).
One may interior if respective product. Could one exceed 1, making endpoint closest. Are distances of two nonincident x edges perhaps differ. Problem states shortest distances ... are three values, singular perhaps six "edges it does not meet" perhaps distances repeat due to parallelepiped symmetry? Wait rectangular parallelepiped maybe "an interior diagonal" could mean a diagonal line segment connecting opposite vertices, and "edges it does not meet":There are? At endpoints it meets? A diagonal meets 6 edges. Remaining 6, but maybe each of 6 pairs via central inversion? Complement maps diagonal endpoints swapped and maps an edge to opposite parallel edge: x edge (y,z)=(b,0) maps under x→a-x etc to x edge (0,c), yes. Distance from same diagonal preserved (central inversion maps diagonal to itself), so distances equal. Segment distance to entire edges must equal due to symmetry. Why foot issue then both must same. If t for one interior and other maybe >1 seemingly impossible symmetry should map. Let's re-evaluate central inversion maps x edge at y=b,z=0 to x edge at y=0,z=c, point parameter x→a-x. Diagonal t→1-t. If closest on first has t0, s0; mapped t1=1-t0. Our formula for second perhaps reciprocal relation but solve gave ac/(b²+c²), inconsistent perhaps distance equations wrong or edge positions / closest.

Let's calculate second P=(0,0,c): difference P+s u - tv. v dot P = c*c = c², not ac! I mistakenly c coordinate dot = c². Orthogonality v: c² + a s - tL²=0, s=at -> t=c²/(L²-a²)=c²/(b²+c²). First t=b²/(b²+c²), not ab. Great. They sum 1. Interior always. Nice.

Similarly formula distance line:
x edge P=(0,b,0), cross u×v = (0,-c,b), dot = -bc => bc/√(...). Other same.

Thus three distinct distances d_x = bc/√(b²+c²), d_y=ac/√(a²+c²), d_z=ab/√(a²+b²).

We need solve with these three given. Let x? Dimensions unknown. Need derive volume. Equations.

We can relate reciprocal squares:
1/d_x² = (b²+c²)/(b²c²)=1/b²+1/c².
Similarly 1/d_y²=1/a²+1/c²; 1/d_z²=1/a²+1/b².
Given rational squares:
d1²=20 => reciprocal =1/20.
d2²=900/13 => reciprocal=13/900.
d3²=225/10=22.5 => reciprocal=2/45 =? 0.044444.Let p=1/a², q=1/b², r=1/c². Three pair sums:
depending assignment:
q+r=1/20 = .05
p+r=13/900≈.014444
p+q=2/45≈.044444
in arbitrary pairing. But the given distances correspond to three unordered edge-directions, so can assign in any order. Pair sums determine p,q,r independent of labeling? Given three sums S1,S2,S3; individual p=(S? two sum minus third)/2. Since labels can map, but all three equations exactly values, solutions are combinations where each of p,q,r equals (sum of two S minus third)/2; set same regardless permutations. So p values:
(S1+S2-S3)/2
(S1+S3-S2)/2
(S2+S3-S1)/2. Need positive.

Compute S_A=1/20=45/900.
S_B=13/900.
S_C=2/45=40/900.
Then:
(A+B-C)=18/900 /2? p = (58-40)/1800=18/1800=1/100.
(A+C-B)=(45+40-13)=72 /1800=1/25.
(B+C-A)=(13+40-45)=8/1800=1/225.
So reciprocal-square dims set {1/100,1/25,1/225}, dimensions {10,5,15}. Volume 750.

Likely answer. But need be meticulous: The "shortest distances between an interior diagonal ... and edges it does not meet" perhaps maybe an "interior diagonal" could be face? Interior diagonal = body diagonal. It does not meet six edges; distances pair equal and formulas. Need perhaps establish closest point lies on segments, not infinite lines. Could show formula and symmetry. Could also perhaps direct coordinate setup.

Need reason step-by-step and final line exact required "Final Answer: <number>" number only no units. We need not put units after. Ensure last line.

Let's fully derive segment distance. Let dimensions x,y,z, diagonal vector (x,y,z), choose O origin. For an edge parallel x not incident, endpoints maybe (0,y,0) to (x,y,0). Need find closest segment points. Line parameter diagonal t(x,y,z), edge (s,y,0), t,s∈[0,1]. Minimize. Supporting lines shortest vector direction cross product. Difference at closest:
? Distance = yz / sqrt(y²+z²). Foot diagonal parameter? derive t=y²/(y²+z²)? Let's verify with edge at y=Y,z=0. As above t=y²/(y²+z²), s=x t. Both in [0,1]. yes. Other nonincident x edge y=0,z=z: t=z²/(y²+z²), also [0,1]. Distances same. Thus valid.

Potential concern rectangular parallelepiped has edges meeting diagonal? Body diagonal endpoints each meets three edges, total six. Remaining six grouped into three pairs. Explain.

Let's derive pair sum assignment. Let
d_x²=y²z²/(y²+z²)
Then 1/d_x²=1/y²+1/z².
Similarly.
Set A=1/x² etc. The three pair sums equal {1/20,13/900,2/45}. Since no labels necessary. Need calculate A,B,C:
Let S_xy? Denote S1,S2,S3.
A=(S_yz? depending). If S_x corresponding y+z. But simply after ordering, solve.
Take
A+B=2/45,
A+C=13/900,
B+C=1/20 maybe choose assignment. Then solve:
Let's map dimensions:
A+B maybe 2/45=40/900
A+C=13/900
B+C=45/900.
Then
A=(first+second-third)/2=(40+13-45)/(1800)=8/1800=1/225 => x=15.
B=(first+third-second)/2=(40+45-13)/1800=72/1800=1/25 => y=5.
C=(second+third-first)/2=(13+45-40)/1800=18/1800=1/100 => z=10.
This assignment yields dimensions 15,5,10. Volume 750. Check corresponding distances:
for x=15,y=5: d_z? xy/sqrt(250)=75/(5√10)=15/√10 yes if A+B sum 2/45 gives d²=22.5. A+C: 15,10: 150/√325 =? 150/(5√13)=30/√13. B+C:5,10=50/√125=2√5. Great.

Need perhaps mention distances to segments and supporting lines coincide due to foots. We can formulate.

But before final, let's examine if thereare hidden ambiguity or alternate interpretation. "The shortest distances between an interior diagonal of a rectangular parallelepiped, P, and the edges it does not meet are ..." It may come from contest where "interior diagonal" likely line segment. The three values perhaps correspond to distances from diagonal to each of six nonincident edges, but due to pairing. Our solution likely 750. Need make no arithmetic mistake.

Let's deeply inspect geometry to ensure formula for shortest segment-line edge and diagonal, particularly if closest points indeed interior. For one family x:
Diagonal Q(t)=(ta,tb,tc), edge E(s)=(s,b,0). Vector E-Q=(s-ta,b-tb,-tc). Derivative conditions if interior:
wrt s zero gives s-ta=0.
wrt t: -a(s-ta)-b(b-tb)+c(tc)=? First zero, -b²+t b²+t c²=0 => t=b²/(b²+c²), yes. s=a t ≤a. Distance at t: coordinates x diff zero, y=b(1-t)=b c²/(sum), z=-c t=-c b²/(sum). Squared = b² c^4/S² + c² b^4/S² = b²c²(c²+b²)/S²=b²c²/S. yes. Other x-edge E(s,0,c): t=c²/(sum), y=-b c²/S? z=c(1-t)=c b²/S, same. Fine.

Could an endpoint give shorter than infinite closest? no because infinite foot in segments. Good.

Can mention central symmetry gives each distance twice.

Now desired answer asks reason step by step. We should provide polished derivation.

Let's consider possible concern: They ask final numerical answer line in format and number only no units. We can have preceding equations. Last line exactly `Final Answer: 750`. No period maybe.

Let's formulate:
Let side lengths a,b,c and coordinate vertices...
Diagonal from (0,0,0) to (a,b,c).
For pair of nonincident edges parallel x, choose one line through (0,b,0). Points:
D(t)=t(a,b,c)
E(s)=(s,b,0)
At perpendicular minimum, maybe use cross product shortcut, but must validate foot in ranges. Infinite-line distance formula:
d_a = |(0,b,0)·((1,0,0)×(a,b,c))| / sqrt(b²+c²)
=bc/sqrt(b²+c²).
Closest D parameter t=b²/(b²+c²) and edge parameter s=at, so within [0,1], so segment distance same. For opposite edge likewise. Then equations.

Define X=1/a² etc. Given:
bc/... distance corresponds one; because labels arbitrary, choose:
1/b²+1/c²=1/20
1/a²+1/c²=13/900
1/a²+1/b²=2/45.
Wait map given values as earlier? We had A+B =2/45 for a=15,b=5; A+C=13/900; B+C=1/20. Let's set equations exactly:
Let A=1/a², B=1/b²,C=1/c².
Could write:
B+C=1/(2√5)^2=1/20.
A+C=1/(30/√13)^2=13/900.
A+B=1/(15/√10)^2=10/225=2/45.
Then solve:
A=(A+B+A+C-(B+C))/2 = (2/45+13/900-1/20)/2. Convert: 40+13-45=8/900 /2 =4/900=1/225 yes.
B=(A+B+B+C-(A+C))/2=(40+45-13)/900 /2=72/1800=1/25.
C=(A+C+B+C-(A+B))/2=(13+45-40)/1800=18/1800=1/100.
Then a=15,b=5,c=10. Volume 750.

Check formula if B+C corresponds edge parallel a = bc distance; yes any assignment. Fine.

We can perhaps avoid labeling which input maps which dimensions because volume invariant, but explicit assignment okay due relabeling.

Need be attentive cross product formula notation: Edge through (0,b,0), direction e_x; diagonal direction v. Shortest line distance absolute `(P · (e_x × v))/|...|`. `e_x×v=(0,-c,b)`, dot with (0,b,0)=-bc. abs. Good. But perhaps deriving foot from formula is extra. We should establish segment min. Could state closest parameters, enough.

Let's maybe provide direct minimization to make robust:
For edge E(s)=(s,b,0), squared distance:
(s-at)^2+(b-bt)^2+(0-ct)^2.Minimizing first in s gives s=at. Then minimize f(t)=b²(1-t)²+c²t². Derivative -2b²(1-t)+2c²t=0 -> t=b²/(b²+c²). Then s=at. Both [0,1]. f_min = b² c^4/S² + c² b^4/S² = b²c²/S. So distance. This is clear and demonstrates segment. The opposite edge gives same. Do that perhaps.

Then sum.

No units in final.

Likely straightforward. But token budget huge; perhaps hidden trick? Let's not rush. Let's inspect wording exactly: "The shortest distances between an interior diagonal of a rectangular parallelepiped , $P$ , and the edges it does not meet are $2\sqrt{5}$ , $\frac{30}{\sqrt{13}}$ , and $\frac{15}{\sqrt{10}}$ . Determine the volume of $P$ ." They list three, confirming pairs. Could "interior diagonal" refer diagonal contained in interior except endpoints, indeed body diagonal.

Let's see if perhaps shortest distance to an edge means shortest distance from *diagonal segment* to edge segment and formulas differ if closest points endpoints. But we showed foots interior for axis-aligned box for each nonincident edge, based diagonal from opposite vertices. Check all edges: For edge parallel a at y=b,z=0 t=b²/(b²+c²), yes. For y=0,z=c t=c²/(...). Both. So yes.

Let's independently verify distances for dimensions 15,5,10 and all six:
Edges parallel a: bc/sqrt(25+100)=50/(5√5)=2√5.
parallel b: ac/sqrt(225+100)=150/(5√13)=30/√13.
parallel c: ab/sqrt(225+25)=75/(5√10)=15/√10.
Matches exact order if dimensions a=15,b=5,c=10. Good.

Could volume perhaps product inverses sqrt formula direct. Our values.

Let's ensure reciprocal of (15/√10)^2:
(15/√10)^2=225/10=45/2=22.5, inverse 2/45. yes.
30/√13 squared 900/13 inverse13/900.
2√5 squared20 inverse1/20.

Solving gives:
A from equations perhaps dimensions aassociated with d_x = B+C=1/20 => 1/a²? We assign:
A+C=13/900
A+B=2/45.
A = [(A+B)+(A+C)-(B+C)]/2 = [40+13-45]/1800=8/1800=1/225 -> a=15.
B = [A+B+B+C-A+C]/2=[40+45-13]/1800=72/1800=1/25 -> b=5.
C=[A+C+B+C-A+B]/2=[13+45-40]/1800=18/1800=1/100 -> c=10.
yes.

Let's craft final response with sections maybe.

Need follow user's final line requirement: "put your final numerical answer on the last line in the format 'Final Answer: <number>' (number only, no units)." We should ensure final line is that. No markdown display after. Could bold? They specify format, use plain.

Let's produce answer around 500 words? Desired oververbosity 5, step-by-step perhaps 300. No need excessively long.

But wait, system says accessed API, no other issue.

Let's think if there might be an alternative issue: A rectangular parallelepiped has 4 body diagonals; any "interior diagonal" maybe diagonals between nonadjacent vertices not on same face, yes all equivalent under box symmetry? A rectangular box's symmetry group can map any body diagonal to any other via reflections at center while dimensions unchanged, so no issue.

Could shortest distances be to "edges it does not meet" where diagonal's line may intersect extensions of some edges, but segment doesn't. We use segments.

Let's perhaps mention diagonal meets six edges: at each endpoint, 3; the other six form 3 opposite pairs, one for each edge direction. Thus exactly three values.

Let's formulate derivation carefully:
Coordinates:
`0≤x≤a,0≤y≤b,0≤z≤c`.
Diagonal from `(0,0,0)` to `(a,b,c)`, parametrized `D(t)=(at,bt,ct)`.
One nonincident edge parallel x is `E(s)=(s,b,0)`. Squared distance. At min wrt s, s=at. Then minimize `g(t)=b²(1-t)²+c²t²`, giving `t=b²/(b²+c²)`, interior, hence `s=at∈[0,a]`. Therefore d_a=bc/sqrt(...). For other x edge perhaps same. Similar.

Then equations.

Let's ensure notation d_a corresponds to edge parallel a-side length, not distance involving a in numerator. Fine.

Potential rounding no.

Let's now prepare final.


Question: Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ?  The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$

For each of the six elements of \(S\), there are three possible membership patterns in the two subsets \(A\) and \(B\):

1. It is in \(A\) only.
2. It is in \(B\) only.
3. It is in both \(A\) and \(B\).

It cannot be in neither subset, since \(A\cup B=S\). Thus, there are

\[
3^6=729
\]

ordered pairs \((A,B)\).

Now account for the fact that order does not matter. The only ordered pair equal to its reverse, meaning \(A=B\), occurs when

\[
A=B=S.
\]

Therefore, by Burnside’s lemma (or by pairing the other \(728\) ordered selections in groups of two), the number of unordered selections is

\[
\frac{729+1}{2}=365.
\]

Final Answer: 365

Explanation:
We need answer stepwise. Need solve. Need be careful unordered pairs subsets union S, allowing equal. n=6. For each element membership pair among in A, in B, both: 3 choices => ordered pairs 3^6=729. Need unordered selections, multisets {A,B}, includes equal. Number unordered pairs = (ordered pairs + number fixed under swap)/2. A=B and A union A=A=S, only one A=S, so (729+1)/2=365. Need explain. Is there possible issue selections "two not necessarily distinct subsets" and order no matter. yes.

Need put final answer last exact. Step by step.


Question: A solitaire game is played as follows.  Six distinct pairs of matched tiles are placed in a bag.  The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand.  The game ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty.  The probability that the bag will be emptied is $p/q,\,$ where $p\,$ and $q\,$ are relatively prime positive integers.  Find $p+q.\,$

Let \(S_m\) be the probability of successfully emptying the bag from a state where:

- \(m\) complete pairs remain untouched in the bag, and
- the player holds exactly one unmatched tile.

Thus \(2m+1\) tiles remain. Clearly \(S_0=1\).

From this state, consider the next draw:

1. **Draw the mate of the held tile.**
   This occurs with probability \(\frac1{2m+1}\). The hand becomes empty, and the next draw necessarily starts a new pair, leaving the same type of state with \(m-1\) untouched pairs.

2. **Draw a tile from an untouched pair.**
   This occurs with probability \(\frac{2m}{2m+1}\). The player then holds two unmatched tiles. Of the remaining \(2m\) tiles, exactly two match a held tile, so the player survives the next draw with probability
   \[
   \frac{2}{2m}=\frac1m.
   \]
   This again leaves the state with \(m-1\) untouched pairs and one unmatched tile.

Therefore,
\[
S_m=\left(\frac1{2m+1}+\frac{2m}{2m+1}\cdot\frac1m\right)S_{m-1}
=\frac{3}{2m+1}S_{m-1}.
\]

After the initial draw, there are five untouched pairs and one unmatched tile, so the desired probability is
\[
S_5=\prod_{m=1}^{5}\frac{3}{2m+1}
=\frac{3^5}{3\cdot5\cdot7\cdot9\cdot11}
=\frac9{385}.
\]

Thus \(p=9\), \(q=385\), and
\[
p+q=394.
\]

Final Answer: 394

Explanation:
We need solve. Need step-by-step final last line exact. Need be careful interpretation. Six distinct pairs = 12 tiles. Draw without replacement. Retain except matching tiles put aside as soon as appear in player's hand. Game ends if player ever holds three tiles no two match. Otherwise until empty. Need probability never state hand has 3 singleton distinct pending tiles; matches clear pair when second drawn.

We need derive. Need perhaps known AIME? Need probability random permutation avoiding 3 distinct unmatched open symbols at any prefix. Need calculate rational reduced p+q.

Let's understand process. At any point, hand contains at most? Singles whose mate not drawn; pairs immediately removed. Failure when 3 open distinct pairs before their partners. So ballot-like with 6 duplicated labels random multiset permutation. Need count permutations where stack/open count never reaches 3. Game may "otherwise drawing until bag empty"; could survive. Starting hand 0, increments first occurrence pair, decrements second. Need paths of 6 opening/closing parentheses with labels, max depth <=2. Count all linear orders of 12 tiles where all pairs distinct and each label two positions. Number equally likely = 12! perhaps labeled. Favor count via states/matchings or Catalan first return structures with pair identities.

Can model sequence of pairs among 6 labels, each twice. Count permutations of 12 distinct physical tiles equivalent each pair's two indistinguishable event positions times 2^6 tile orientations. Probability count event words over labels each twice / multinomial 12!/(2!)^6, same. Need count depth <=2. Could dynamic program states based number unmatched n=1,2 perhaps and labels.

There may be formulas. We need exact fraction. Let's derive multiple ways verify.

At a given number k remaining? Transitions:
state i currently unmatched (0,1,2), m pairs? Need distinguish counts of untouched pairs and open pairs. Draw probability next opens an untouched pair or closes an open pair. If all tile identities equally random.
Let state (u,o) where u = number pairs neither tile drawn, o = number unmatched in hand, total remaining tiles 2u+o. Next draw: open: probability 2u/(2u+o) choosing one tile from untouched pair; close: o/(...). Goal paths from (6,0) to (0,0), avoiding transition open at o=2 (would hold 3). Since monotonic dimensions. Probability product.

Couldsum probabilities over lattice paths. State dimensions u+o <=6. Dynamic calculate rational. Easier count weighted paths or recursive probabilities.

At state (u,o), if opening increases o to 3 at o=2 -> failure. Closing decrements o. Need probability eventually all close with open choices only until o<=1. This is essentially queue capacity 2. We can calculate dynamic recursion P(u,o):
P(0,0)=1; for o=0,1,2 allowed:
P(u,o)= [2u/(2u+o)] P(u-1,o+1) + [o/(2u+o)] P(u,o-1), with opening term absent if o=2. Evaluate P(6,0). Rational perhaps sizeable. Let's do exact carefully.

Alternative combinatorial count all words max stack 2. Number perhaps use formula involving Catalan? For n pairs and capacity 2, number words =? Event label sequences. Let's inspect. There are n distinct labels, each twice, and stack depth <=2. This is words avoiding pattern? Count via DP can produce integer and denominator total (2n)!/2^n. Let's derive to cross-check.

At state (u,o), event sequence future count F(u,o) (# arrangements of remaining event tokens: u pairs both unseen + o single closers, maintaining capacity). Recurrence F(u,o)=F(u-1,o+1)*? Number choices label opening among u, and physical tile issue if event sequence; in distinct physical remaining perhaps 2u choices, o choices. Event words recurrence u F(u-1,o+1)+ F(u,o-1). Total words event = (2n)!/2^n. Physical=2^n event= (2n)!? Wait each event word corresponds 2^n physical orderings (within each matched pair), so denominator physical 12!, numerator physical=2^6 * event count. Ratio event favorable / total event words. Recurrence event:
G(u,o)=u G(u-1,o+1)+G(u,o-1), boundary G(0,0)=1, no open at o=2. Compute. Then probability G(6,0) / (12!/2^6). Could be integer manageable.

Let's compute either. Let's dynamic exact probabilities perhaps numerator denominator after reduction. Need final p+q likely AIME <=? p/q maybe denominator reduced and sum perhaps e.g. 141? Let's calculate.

We can use recursive weighted path. Let'screate tables for G.

State capacity 2. For o=0: G(u,0)=u G(u-1,1) (must open).
o=1: G(u,1)=u G(u-1,2)+G(u,0).
o=2: G(u,2)=G(u,1) (must close one of two; event recurrence coefficient one; physical coefficient 2).
Base G(0,0)=1; G(0,1)=G(0,0)=1; G(0,2)=1.

Compute:
u=0: G00=1,G01=1,G02=1.
u=1:
G10=1*G01=1.
G11=1*G02 + G10=1+1=2.
G12=G11=2.
u=2:
G20=2 G11=4.
G21=2 G12 + G20=4+4=8.
G22=8.
u=3:
G30=3 G21=24.
G31=3 G22 + G30=24+24=48.
G32=48.
Pattern G(u,2)=G(u,1); maybe. Recurrences:
G_{u,2}=G_{u,1}
G_{u,1}=u G_{u-1,2}+G_{u,0}
G_{u,0}=u G_{u-1,1}.
Thus G_u1 = u G_{u-1,1}+u G_{u-1,1}=2u G_{u-1,1}. With G0,1=1 -> G_u1 = 2^u u!. Then G_u0=u G_{u-1,1}=u * 2^{u-1}(u-1)!=2^{u-1}u!. For u=6: 2^5*720=23040. Is that all? Seems pattern yes. Total event words 12!/2^6 = 479001600/64 = 7,484,400. Favor 23,040. Ratio = 23040 / 7484400. Simplify. 7,484,400 / 23,040 =? 324.84375. Factor:
23,040 = 2^? 23040 = 2304*10 =2^8*9 *2*5 =2^9*3^2*5 =512*45 yes.
Total event = 12!/64.12! factor: 479001600 = 2^10 *3^5*5^2*7*11? Let's factor 12!: floor 2=6+3+1=10; 3=4+1=5; 5=2;7,11. /2^6 =>2^4*3^5*5^2*7*11. Numerator 2^9*3^2*5. But numerator has higher 2 exponent than denominator due event counts. gcd = 2^4*3^2*5=16*45=720. Ratio numerator/720=32 (2^5); denominator/720: total 7,484,400/720 =? 10,395. Factor 3^3*5*7*11 =27*385=10395. So p=32 q=10395, sum 10427. Is that answer likely? Let's sanity.

But wait our recurrence may have undercounted physical arrangements? Probability ratio event words yes all event words equally likely? Random draw distinct tiles: each event pattern (sequence labels ignoring which copy) probability? Each event pattern with each label twice has 2^6 physical tile assignments? Given event label sequence, for each label, first and second positions can be assigned tile A/B in 2 ways, independently, yes 2^6. Thus equal. Good.

Yet let's examine process phrase "retains them, except that matching tiles are put aside as soon as they appear in player's hand." If player draws matching tile to a retained tile, both put aside. Ends if player ever holds three tiles, no two match. Our stack depth. Fine.

But perhaps game ends as soon as hand has three unmatched after drawing; yes.

Could answer 10427. Need give step-by-step. Let's ensure recurrence count logic and total sample. We can offer elegant state count.

Let's fully reason maybe define N(u,o) valid ways to complete from state with u unstarted pairs and o unmatched tiles currently held. However labels: u distinct untouched pairs and o distinct singles. Number of future event orderings where label appears etc. At state:
- If o<2, choosing next tile that starts one of u unopened pairs. In event sequence, u choices for its label; state (u-1,o+1).
- If o>0, choosing next tile matching one of o held singleton labels: o? Wait event word recurrence I wrote + F(u,o-1)without coefficient o. Why? Event sequences are sequences of labels, and current open labels are distinct; choosing which open label to close gives o choices! I mistakenly omitted o multiplier! Huge issue. Need revisit. In event word recurrence, opening label u choices; closing label o choices. I wrote G(u,o)=u G(u-1,o+1)+G(u,o-1), but should be o G(u,o-1). In physical tile count recurrence closing has o choices (only one counterpart tile each), opening has 2u choices. Event recurrence after divide? Let's be exact. I conflated. Need recalc. Probability recursion had closing probability o/(2u+o), opening 2u; so event-weight recurrence should u? Let's see event count total state words labels each occurrences: for untouched u, each twice; open o labels each once. Number = (2u+o)!/2^u. Recurrence should u F(u-1,o+1) + o F(u,o-1). Check total recurrence T(u,o)=(2u+o)!/2^u. u T(u-1,o+1)=u(2u+o-1)!/2^{u-1}=2u factorial term; o T(u,o-1)=o factorial term; sum (2u+o)*(...)= yes. Thus closing multiplier o. I omitted. Favorable much lower. Need compute carefully.

Recurrence:
F(0,0)=1 perhaps F(0,o)=o! (close open labels in any order) for o<=2.
At capacity, o=2 cannot open; F(u,2)=2 F(u,1) (choose one of 2 open).
o=1: F(u,1)=u F(u-1,2)+1 F(u,0).
o=0: F(u,0)=u F(u-1,1).
Compute.

Could derive table. Let's calculate exact. Let A_u=F(u,0), B_u=F(u,1), C_u=F(u,2).
Base u=0: A0=1, B0=1, C0=2.
Recurrences u>=1:
A_u = u B_{u-1}
B_u = u C_{u-1} + A_u
C_u = 2 B_u.
Then substitute:
C_{u-1}=2 B_{u-1}.
B_u=2u B_{u-1}+u B_{u-1}=3u B_{u-1}. Ah because A_u=u B_{u-1}. So B_u=3u B_{u-1}, B0=1 => B_u=3^u u!. Then A_u=u*3^{u-1}(u-1)!=3^{u-1}u!. C=2*3^u u!. For n=6 A=3^5*720=243*720=174,960. Total event words=7,484,400. Ratio 174960 / 7484400. Simplify. Total factor 2^4*3^5*5^2*7*11. Favorable 3^5 * 6! =3^5*(2^4*3^2*5)=2^4*3^7*5. gcd 2^4*3^5*5. Denominator left 5*7*11=385. Numerator left 3^2=9. Probability=9/385. p+q=394. Is this likely? Let's test via probability recursion product maybe. This seems more plausible.

Let's simulate intuitive probability survive 6 pairs capacity 2 ~0.0234 (9/385). Could be.

Let's calculate direct dynamic probabilities to verify:
At each open state:
P(u,0)=P(u-1,1) (must open; transition certain though state u decreases).
P(u,1)= [2u/(2u+1)] P(u-1,2)+[1/(2u+1)]P(u,0).
P(u,2)=P(u,1) because must close one, transition certain to u,1 (regardless which). Note P(u,2)=P(u,1).
Then P(u,0)=P(u-1,1).
P(u,1)= [2u/(2u+1)]P(u-1,1)+[1/(2u+1)]P(u-1,1)=P(u-1,1)! Wait then all 1? Something wrong: At o=2, closing transition is certain (draw one of held counterpart; indeed any next tile? If hand holds 2 unmatched, bag has untouched tiles too; drawing an untouched tile would make 3 and game ends. The transition is not certain success; if draw close, survive to o=1; if open, fail. In recursion P(u,2)= [o/(2u+o)]P(u,o-1), not simply P(u,1), because opening outcomes fail. In counting F(u,2)=o F(u,o-1) coefficient 2. Right. Then:
P(u,2)=2/(2u+2)=1/(u+1) P(u,1).
P(u,0)=1 * P(u-1,1) (draw untouched guaranteed when hand empty; yes)
P(u,1)=2u/(2u+1) P(u-1,2)+1/(2u+1)P(u,0)
=2u/(2u+1)*(1/u) P(u-1,1)? P(u-1,2)=1/u P(u-1,1). First =2/(2u+1) Pprev. Second=1/(2u+1) Pprev. total=3/(2u+1)Pprev. Starting P(0,1)? At u=0,o=1 only close certain =>1. So P(n,0)= product u=1..n? P(u,1)=3/(2u+1) P(u-1,1), and P(u,0)=P(u-1,1). At P(6,0)=P(5,1)= product_{u=1}^5 3/(2u+1)=3^5/(3*5*7*9*11)=? 243/(10395)=? Simplify 81/3465=27/1155=9/385. Yes. Note recurrence count A_6 corresponds probability maybe A_6 / total event =9/385. Good.

Wait indexing: P(u,o) success from state. Start (6,0). P(6,0)=P(5,1), product states u=1..5 as above. Counting A_6=3^{5}6!, denominator total 12!/2^6, ratio:
3^5*720 / 7484400 = 174960/7484400 = divide? 19440? 9/385 yes. Good.

Let's thoroughly ensure event state recurrence coefficients. Could present perhaps simpler product after drawing first tile. The game must proceed through a pattern: Starting empty, first draw necessarily opens. Then success imposes something. We can derive product 3/(2m+1), butneed ensure product index and rationale transparent.

Let's identify state after initial draw: 5 untouched pairs and one singleton. More generally, define S_m = probability of successfully emptying from a state with m unopened pairs and one unmatched tile, given we are safe. Perhaps derive recurrence elegantly. At state m≥1:
- one unmatched; 2m untouched tiles, total 2m+1.
Next draw:
  * probability 1/(2m+1): draw mate, hand becomes empty with m pairs unopened. Then next draw necessarily opens one, transitioning to state m-1 with one unmatched. This contributes S_{m-1}.
  * probability 2m/(2m+1): draw from unopened pair, hand has two unmatched. To avoid immediate loss, the next draw must be one of their two mates among total 2m tiles remaining? Let's count after opening one: untouched pairs m-1 =2m-2 tiles, plus two open mates =2m total. Probability immediate close =2/(2m)=1/m. Then state returns to one unmatched (the first singleton) and m-1 unopened pairs. Thus contribution [2m/(2m+1)]*(1/m) S_{m-1}.
Total S_m = [1/(2m+1)+2/(2m+1)] S_{m-1}=3/(2m+1) S_{m-1}. Base S_0=1. Starting after first tile S_5. Product m=1..5 3/(2m+1)=3^5/(3*5*7*9*11)=? 243/10395=9/385. Very neat. Need be careful after matching the initial singleton: hand empty, next draw from m unopened pairs necessarily gives one unmatched, so yes S_{m-1}. No probability factor (1) and cannot lose on that draw because only one held. In second branch after two unmatched, next draw must close one; if not lose. If closes, which one irrelevant; state m-1 untouched and one unmatched. Correct.

But potential issue: If at one unmatched state and draw an unopened tile, hand has 2; next draw closing gives one unmatched. Fine. If next draw matches just-opened or older singleton, either one remains, same state. Number of successful closing tiles 2. Good.

Base S_0=1: with only one unmatched? State m=0and one unmatched: next draw necessarily its mate and empty, success. So 1.

Then starting first draw from 12 always leaves one unmatched and 5 untouched pairs, so S5.

Calculate:
S5 = ∏_{m=1}^5 3/(2m+1)
=3^5/(3·5·7·9·11)
=243/10395. Divide by 27? 243/10395: 10395 /27=385, 243/27=9. Yes 9/385. gcd 9,385? 385=5*7*11, no 3. p+q=394.

Need provide step-by-step and final line exact "Final Answer: 394". User specifically final numerical answer last line format. We can give explanation then last line. Ensure no text after. No units.

Let's make answer desired oververbosity 5: state definition, transitions, product, reduction. Need ensure interpretation clear.

Let's perhaps validate no subtle distinction "game ends if player ever holds three tiles, no two of which match." Could potentially holding 3 where two match impossible because matches immediately put aside, so any three held necessarily no two match. Fine.

Let's formulate:

Let \(S_m\) denote probability of eventually emptying bag from safe position in which:
- m complete untouched pairs remain,
- exactly one unmatched tile is in hand.
(All previous completed pairs gone.)

For m=0, S0=1.

For m>=1, 2m+1 tiles remain (m pairs plus mate of held tile). On next draw:
1. mate probability 1/(2m+1): hand empty, m pairs remain. Following draw (certain to be safe) starts one pair, yielding state parameter m-1, one unmatched. Contribution etc.
2. one of 2m tiles from untouched pairs probability 2m/(...). Then hand has 2 unmatched and 2m tiles remain: 2 are mates, 2m-2 untouched. To survive immediate next draw, must choose mate, probability 2/(2m)=1/m. Then state m-1, one unmatched.
Thus S_m = [1/(2m+1)]S_{m-1} + [2m/(2m+1)](1/m)S_{m-1} =3/(2m+1) S_{m-1}.

Actually in first case after mate draw, next tile draws from 2m tiles all untouched.Could it ever match something? hand empty, no. state m-1. Certain. Good.

Initial draw: certain leaves five untouched pairs, one singleton. So probability S5. Product. Show arithmetic.

One nuance: In branch 2 after drawing from untouched pair, total remaining is 2m? Initially at state m, total 2m+1; after one draw, 2m. Two are mates of current two held tiles. yes.

p=9,q=385 sum 394.

Let's not overcomplicate with combinatorial count. Ensure final line no dollar perhaps exactly. They request number only no units within format. We can use Final Answer: 394.

Let's produce.


Question: For each real number $x$ , let $\lfloor x \rfloor$ denote the greatest integer that does not exceed $x$ . For how many positive integers $n$ is it true that $n<1000$ and that $\lfloor \log_{2} n \rfloor$ is a positive even integer?

For \(\lfloor \log_2 n\rfloor=k\), we must have
\[
k\le \log_2 n<k+1,
\]
or equivalently,
\[
2^k\le n<2^{k+1}.
\]

Since \(n<1000<1024=2^{10}\), the possible positive even values of \(k\) are
\[
2,4,6,8.
\]

For each \(k\), the interval \([2^k,2^{k+1})\) contains \(2^k\) integers. Thus the total is
\[
2^2+2^4+2^6+2^8
=4+16+64+256
=340.
\]

Final Answer: 340

Explanation: We need solve. Need reason step by step final exact format. Need count positive integers n<1000 such that floor log2 n positive even integer. Let's inspect. floor log2 n = k, positive even k perhaps 2,4,6,8,10? Since n<1000, log2 n < ~9.966. floor max 9. Positive even floors 2,4,6,8. For n positive integer and n<1000. Conditions floor = k means 2^k <= n < 2^(k+1), integer n. Count 2^k values (n=2^k through 2^(k+1)-1 inclusive). Sum k=2,4,6,8: 4+16+64+256=340. But check n<1000 excludes n up to 511 anyway, all below. k=10 requires n>=1024, no. n=1 floor=0 not positive. Count 340. Need perhaps articulate stepwise. Ensure "positive even integer" no ambiguity. Could count intervals. Let's produce concise step-by-step and final last line exact.


Question: The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles, of which $s$ are squares.  The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n.$

There are 9 horizontal and 9 vertical grid lines.

- A rectangle is determined by choosing 2 horizontal lines and 2 vertical lines:
\[
r=\binom{9}{2}\binom{9}{2}=36^2=1296.
\]

- For squares of side length \(k\), where \(1\le k\le 8\), there are \((9-k)^2\) possible squares. Thus
\[
s=\sum_{k=1}^{8}(9-k)^2
=\sum_{j=1}^{8}j^2
=\frac{8\cdot9\cdot17}{6}
=204.
\]

Therefore,
\[
\frac{s}{r}=\frac{204}{1296}=\frac{17}{108}.
\]
Since \(17\) and \(108\) are relatively prime,
\[
m+n=17+108=125.
\]

Final Answer: 125

Explanation:
We need solve. Need step-by-step, final last exact. Need be careful interpretation 9 lines each grid, all axis-aligned rectangles. r choose horizontal pair and vertical pair = C9,2^2=36^2=1296. Squares in 8x8 checkerboard axis-aligned, sum_{k=1}^8 (9-k)^2 = 8^2+...+1 = 204. s/r=204/1296 simplify gcd? 204 factors 2^2*3*17.1296=36^2=2^4*3^4? 36=2^2*3^2, squared 2^4 3^4=16*81. gcd 4*3=12. 17/108. m+n=125. Need explain perhaps count. Ensure "rectangles" could count square included yes. Nine horiz and vert. We need stepwise. Final line exact. No units. Desired oververbosity 5 but straightforward. Need make sure numerator denominator relatively prime. Let's formulate.

Counting rectangles choose 2 horizontal and 2 vertical. r = 36*36 1296.

For square side length k (in unit-cell widths), available horizontal and vertical positions each 9-k, so count (9-k)^2. Sum k1 through8 = sum j1 through8 j² = 8*9*17/6 = 204. Then ratio 204/1296=17/108. sum125.

Check arithmetic formula 8*9*17=1224/6=204 yes.

Last line exactly requested.


Question: Let $u$ and $v$ be integers satisfying $0 < v < u$ . Let $A = (u,v)$ , let $B$ be the reflection of $A$ across the line $y = x$ , let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$ . Find $u + v$ .

The reflections give the coordinates
\[
A=(u,v),\quad B=(v,u),\quad C=(-v,u),\quad D=(-v,-u),\quad E=(v,-u).
\]

The points \(B,C,D,E\) form a rectangle with side lengths \(2v\) and \(2u\), so its area is
\[
(2v)(2u)=4uv.
\]

The remaining portion of the pentagon is triangle \(ABE\). Its base \(BE\) has length \(2u\), and its perpendicular height is \(u-v\). Thus,
\[
[ABE]=\frac12(2u)(u-v)=u(u-v).
\]

Therefore, the area of \(ABCDE\) is
\[
4uv+u(u-v)=u^2+3uv=u(u+3v).
\]
Given that this equals \(451\),
\[
u(u+3v)=451=11\cdot 41.
\]

Since \(u+3v>u\), the possible factor pair is
\[
u=11,\qquad u+3v=41.
\]
Hence
\[
3v=30\implies v=10,
\]
which satisfies \(0<v<u\). Therefore,
\[
u+v=11+10=21.
\]

Final Answer: 21

Explanation:
We need solve. Need carefully reason order/pentagon perhaps simple? Given integers 0<v<u. Coordinates:
A=(u,v)
B reflection across y=x => (v,u)
C reflect B y-axis => (-v,u)
D reflect C x-axis => (-v,-u)
E reflect D y-axis => (v,-u)
Pentagon A B C D E. Need area =451. Determine u+v.

Compute polygon area shoelace likely perhaps area formula? Points symmetric-ish. Let's list:
A (u,v)
B (v,u)
C (-v,u)
D (-v,-u)
E (v,-u)
back A.
This is rectangle [-v,v] x [-u,u] with? Polygon traces A on lower/right-ish near right x u > v. Sequence: A to B diagonal; B->C top horizontal; C->D left vertical; D->E bottom horizontal; E->A right? E=(v,-u) to A=(u,v) diagonal. Shape is rectangle width 2v height 2u =4uv, but two corners? Sequence A/B on right/top and E/A right/bottom. Specifically rectangle corners perhaps B? top edge C-B, bottom D-E; right rectangle corner (v,u)=B and (v,-u)=E are included, while A extends right, forming triangular protrusion replacing right vertical between B,E? Polygon path B->...->E->A->B. Thus area rectangle 4uv plus triangle ABE base 2u and horizontal altitude u-v: area u(u-v). Or perhaps path orientation? A->B->C->D->E then A clockwise likely. Formula area = 4uv + u(u-v)=u^2+3uv = u(u+3v). Check shoelace. Need 451. Since u,v positive ints. 451 factorization = 11*41? 451 yes 11*41. u(u+3v)=451. u integer factor of 451, u positive. Factor candidates 1,11,41,451. Since u>v and positive. u^2 < u(u+3v)=451, so u < sqrt451~21.2 => u 1 or 11. u=1 impossible v<1 and >0 integer. u=11 gives 11+3v=41 => v=10. sum21. Likely answer.

Need reason step by step. But let's scrutinize potential issue: Area pentagon maybe intersections? Sequence forms simple convex? A (u,v) between B and E x etc. Polygon A-B-C-D-E. Convex? For u>v, A protrudes right, all points: A(u,v), B(v,u), C(-v,u), D(-v,-u), E(v,-u). A is likely vertex if v<u, yes. Shape convex? Edge D-E horiz; E-A vector (u-v,u+v); A-B (v-u,u-v)=(-d,d). Cross perhaps all clockwise positive/negative. likely. area formula.

Could use shoelace. They ask step-by-step. Need final line exact format "Final Answer: 21" number only? They say final numerical answer last line format 'Final Answer: <number>' (number only, no units). Interpret number only after colon. We need produce reasoning.

Let's derive area. Reflections coordinates accurately:
A=(u,v).
Reflect over y=x: B=(v,u).
Reflect over y-axis: C=(-v,u).
Reflect over x-axis: D=(-v,-u).
Reflect y-axis: E=(v,-u).

Area decomposition: rectangle with x from -v to v, y from -u to u is bounded by x=±v,y=±u. The polygon includes B,C,D,E and triangle ABE to right of x=v. Does it fully include rectangle? Polygon edge B-C top, C-D left, D-E bottom, and E-A-B encloses triangle to right. Yes union rectangle + right triangle. Rectangle area (2v)(2u)=4uv. Triangle ABE vertical base BE length 2u, distance from A to line x=v is u-v, area 1/2*(2u)(u-v)=u(u-v). Total 4uv+u^2-uv=u^2+3uv.

Then equation. Since gcd perhaps solve. u(u+3v)=451=11*41. Could note u+3v > u. Positive factor pairs (1,451),(11,41). Also factor pair? 451 only. Need ensure u factor. Since product. u could be divisor. Given v positive and v<u:
u+3v between u and 4u, but factor pairs.
Because 1 impossible u>0? if u=1 no integer v satisfying. Or product factor pair. Could present u divides 451, and u^2<451, so u ≤? Positive divisor under sqrt: 1 or11. u=1 impossible due 0<v<u; u=11. Then v=10. Sum.

Could alternatively factor pair direct: u(u+3v)=451, with second factor > first. Positive factor pairs (1,451), (11,41). u=1 impossible; so. Need perhaps 451 factorization proof.

Check whether factor pair (41,11) violates second > first. (451,1). Fine.

We should not overdo. Last line exactly. Ensure no content after. Let's formulate in desired oververbosity 5, stepwise.

Potential concern: If "number only, no units" means final answer line says Final Answer: <number>, yes.

Let's answer.


Question: A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Let \(p_n\) be the probability that the bug is at its starting vertex after \(n\) moves.

To be at the starting vertex after move \(n\), the bug must be at one of the other two vertices after \(n-1\) moves, then select the starting vertex with probability \(1/2\). Thus,

\[
p_n=\frac{1-p_{n-1}}{2}, \qquad p_0=1.
\]

The fixed point is \(1/3\), so

\[
p_n-\frac13=-\frac12\left(p_{n-1}-\frac13\right).
\]

Therefore,

\[
p_n=\frac13+\frac23\left(-\frac12\right)^n.
\]

For \(n=10\),

\[
p_{10}=\frac13+\frac23\left(\frac1{1024}\right)
=\frac{171}{512}.
\]

Hence \(m=171\), \(n=512\), and

\[
m+n=171+512=683.
\]

Final Answer: 683

Explanation:
We need answer step by step. Need parse ambiguity: "probability that the bug moves to its starting vertex on its tenth move" Means lands there on move 10, regardless prior. Random walk triangle transition to either other equally likely. Starting A. Probability at A after 10 = recurrence p_n = probability previous not A * 1/2 = (1-p_{n-1})/2. p0=1. Solve. p_n perhaps 1/3 + (2/3)(-1/2)^n. For n10 = 1/3 + 2/3 * 1/1024 = 1/3+1/1536=512/1536+1/1536=513/1536=171/512. sum 683. Need perhaps wording "moves to its starting vertex on its tenth move" could mean first time? Usually AIME problem maybe "on its tenth move" likely lands. Given probability m/n answer likely 683. Let's ensure recurrence. p10 yes even cycles possible. Triangle complete graph K3 no loops. Return after n = [2+(-1)^n?] Standard p = 1/3 +(2/3)(-1/2)^n. n10=171/512. m+n=683.

Need step-by-step concise. They specifically request reason. Explain symmetry maybe let p_n. Need be cautious: on each move picks one of two vertices where not current. To be at starting A after move n, previous at either B/C, each selects A 1/2. Recurrence.

Derive closed form:
p_n=(1-p_{n-1})/2
p_n - 1/3 = -1/2(p_{n-1}-1/3)
p0=1 => p10=1/3+(2/3)(1/1024)=? Compute common:
2/3*1/1024=2/3072=1/1536. 1/3=512/1536. total 513/1536 reduce /3=171/512. m+n 683.

Could alternatively count paths. 2^10 equiprobable move sequences. Return sequences count? 342 perhaps p 342/1024=171/512. m+n 683. Check count formula. Fine.

Final last exact format number only after phrase. Ensure final numerical answer last line.


Question: In order to complete a large job, 1000 workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then 100 workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional 100 workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the 800 workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?

Let one scheduled quarter of the project require \(W\) work and take \(T\) time with all 1000 workers.

The schedule allows \(4T\) total time. Since the amount of work is fixed, completion time is inversely proportional to the number of workers.

1. **First quarter**
\[
T_1=T
\]

2. **Second quarter with 900 workers**
\[
T_2=T\left(\frac{1000}{900}\right)=\frac{10T}{9}
\]

3. **Third quarter with 800 workers**
\[
T_3=T\left(\frac{1000}{800}\right)=\frac{5T}{4}
\]

Thus, the time used after three quarters is
\[
T+\frac{10T}{9}+\frac{5T}{4}
=\frac{121T}{36}.
\]

The time remaining on the original schedule is
\[
4T-\frac{121T}{36}=\frac{23T}{36}.
\]

Let \(x\) be the number of additional workers hired. With \(800+x\) workers, the final quarter requires
\[
T\left(\frac{1000}{800+x}\right)
\]
time. To finish on schedule:
\[
\frac{1000}{800+x}T\leq \frac{23T}{36}.
\]

Therefore,
\[
800+x\geq \frac{36000}{23},
\]
so
\[
x\geq \frac{36000}{23}-800
=\frac{17600}{23}
\approx 765.22.
\]

The minimum whole number of additional workers is therefore:

Final Answer: 766

Explanation:
We need solve. Need carefully parse ambiguity quarters perhaps quarter of total work? Wording first quarter work, then second quarter etc likely quarters by work, each equal amount, with varying workers. 1000 scheduled total. Need catch up after 3/4, 800 remain? Wait laid off 100 then additional 100 => 800 yes. Need minimum add after third quarter so entire project by original schedule. Establish scheduled durations under 1000: each quarter takes T, total 4T. First quarter with 1000 on schedule T. Second with 900 takes 1000/900 T = 10/9 T, behind 1/9 T. Third with 800 takes 10/8 T=5/4 T. Elapsed: 1 + 10/9 + 5/4 = common36:36+40+45=121/36 T =3.3611 T. Remaining scheduled time =4 -121/36=23/36 T (~0.6389). Final quarter workload takes 1000 workers T, or with 800+x duration 1000/(800+x) T. Need <=23/36. x >= 1000*36/23 -800 =36000/23 -18400/23=17600/23 ≈765.217 => integer min 766. Final.

But inspect wording "just enough complete on schedule", "first quarter ... then ... second quarter ... third..." Could perhaps quarters refer time periods schedule? Wait first quarter of work explicitly. If all 1000 first quarter completed schedule. Then worker reductions cause same "second quarter of work" longer. Classic work problem, answer likely 766. Need step-by-step. They ask minimum number additional workers beyond 800 hired after three quarters completed so project schedule/before. Need explain units worker-days and schedule quarters.

Could there be nuanced continuous hiring immediately after third quarter. Let worker rate r, one scheduled quarter work W. 1000 r * T = W. Let schedule total 4T. Actual first durations. Calculate.

Need final last line exact format number only after colon. Ensure no content after. Could mention fractional capacity and ceil.

Let's formulate moderately detailed. Check 17600/23: 23*765=17595, remainder5 =>765.217, ceil766. With 1566 workers, final duration 1000/1566 T = 500/783 ≈.63857 <=23/36=.63889. With 765 =>1565, duration 200/313≈.638977 >23/36. Fine.

Let's final.