Small Hint:

Substitute $t=\sqrt[4]{x}$

Big Hint:

After clearing the denominator, factor the resulting quadratic in $t$

Solution:

Let $t=\sqrt[4]{x},$ so $t\geq0.$ The equation becomes $$t(7-t)=12,$$ or $t^2-7t+12=0.$ Thus $t=3$ or $t=4,$ and both values are valid in the original equation. Hence $x=3^4=81$ or $x=4^4=256,$ and their sum is $81+256=337.$

Small Hint:

Pair factors so that each pair is a difference of squares

Big Hint:

After the first pairing, the remaining radicals occur only through $\sqrt{42}$

Solution:

Pair the first two factors, then the last two: $$\begin{gathered} (\sqrt6+\sqrt7)^2-(\sqrt5)^2\\ {}=8+2\sqrt{42},\\ (\sqrt5)^2-(\sqrt7-\sqrt6)^2\\ {}=-8+2\sqrt{42}. \end{gathered}$$ Therefore the requested product is $$\begin{aligned} &(8+2\sqrt{42})(-8+2\sqrt{42})\\ &\qquad=168-64=104. \end{aligned}$$

Small Hint:

Express the sum of the cotangents using $\tan x$ and $\tan y$

Big Hint:

Use the tangent addition formula after finding $\tan x\tan y$

Solution:

Since $$\cot x+\cot y =\frac{\tan x+\tan y}{\tan x\tan y},$$ the given equations yield $\tan x\tan y=\frac{25}{30}=\frac{5}{6}.$ Therefore $$\begin{aligned} \tan(x+y) &=\frac{\tan x+\tan y} {1-\tan x\tan y}\\ &=\frac{25}{1-\frac56}\\ &=150. \end{aligned}$$

Small Hint:

Let $S=x_1+x_2+x_3+x_4+x_5$

Big Hint:

Each equation has the form $S+x_i=$ a constant

Solution:

Put $S=x_1+x_2+x_3+x_4+x_5.$ The five equations say that $S+x_i$ equals $6,$ $12,$ $24,$ $48,$ $96,$ respectively. Adding them gives $$\begin{aligned} 6S&=6+12+24+48+96\\ &=186, \end{aligned}$$ so $S=31.$ Hence $x_4=48-31=17$ and $x_5=96-31=65.$ The requested value is $3(17)+2(65)=181.$

Small Hint:

Reduce $n^3+100$ modulo $n+10$

Big Hint:

The condition makes $n+10$ a positive divisor of a fixed integer

Solution:

Modulo $n+10,$ we have $n\equiv-10.$ Thus $$\begin{gathered} n^3+100 \equiv(-10)^3+100\\ \equiv-900\pmod{n+10}. \end{gathered}$$ The condition is therefore equivalent to $900$ being divisible by $n+10.$ Since $n$ is positive, $n+10$ is a positive divisor of $900,$ and its largest possible value is $900.$ This gives $n=900-10=890.$

Small Hint:

Compare $1986$ with consecutive triangular numbers

Big Hint:

The excess over $1+2+\cdots+n$ is the repeated page number

Solution:

Consecutive triangular numbers around $1986$ are $$\begin{aligned} \frac{62\cdot63}{2}&=1953,\\ \frac{63\cdot64}{2}&=2016. \end{aligned}$$ Hence the book has $62$ pages, and the extra summand is $1986-1953=33,$ which is indeed a valid page number.

Small Hint:

These are precisely the numbers whose base-$3$ digits are all $0$ or $1$

Big Hint:

Write the index in base $2,$ then reinterpret those same digits in base $3$

Solution:

A sum of distinct powers of $3$ has only $0$’s and $1$’s in base $3.$ As these strings increase, they occur in the same order as binary numerals with the same digit strings. Since $$100=1100100_2,$$ the $100$th positive term is $$\begin{aligned} 1100100_3 &=3^6+3^5+3^2\\ &=729+243+9\\ &=981. \end{aligned}$$

Small Hint:

Factor $1000000$ and count all of its positive divisors

Big Hint:

Pair every divisor $d$ with its complementary divisor $\frac{1000000}{d}$

Solution:

Let $N=1000000=2^6 5^6.$ It has $(6+1)(6+1)=49$ positive divisors. The product of all of them is $N^{\frac{49}{2}},$ so the sum of their base-$10$ logarithms is $$\frac{49}{2}\log_{10}N=\frac{49}{2}\cdot6=147.$$ The proper divisors include $1$ but exclude $N$ itself. Subtracting $\log_{10}N=6$ gives $S=141,$ already an integer.

Small Hint:

Normalize the perpendicular distances from $P$ to the three sides

Big Hint:

A cross-section parallel to a side has length equal to that side times one minus the corresponding normalized distance

Solution:

Write $a=BC=450,$ $b=CA=510,$ and $c=AB=425.$ Let $x,$ $y,$ and $z$ be the distances from $P$ to $BC,$ $CA,$ and $AB,$ respectively, each divided by the corresponding altitude. Area decomposition gives $x+y+z=1.$

By similar triangles, the segment through $P$ parallel to $BC$ has length $a(1-x),$ and similarly the other two lengths are $b(1-y)$ and $c(1-z).$ Since all three equal $d,$ $$\begin{aligned} x&=1-\frac da,\\ y&=1-\frac db,\\ z&=1-\frac dc. \end{aligned}$$ Their sum is $1,$ so $$d=\frac{2abc}{ab+bc+ca}.$$ Substituting the three side lengths gives $$\begin{aligned} d&=\frac{195075000}{637500}\\ &=306. \end{aligned}$$

Small Hint:

First include the original number and sum all six permutations

Big Hint:

If $s=a+b+c,$ express the original number in terms of $s$ and $N$

Solution:

Across all six permutations, each digit occurs twice in each place, so their total is $222(a+b+c).$ Put $s=a+b+c$ and let the original number be $M.$ Since the other five sum to $3194,$ $$M=222s-3194.$$ Because $100\leq M\leq999,$ we need $15\leq s\leq18.$ Testing these four values gives $M=136,$ $M=358,$ $M=580,$ and $M=802,$ respectively. Only $358$ has digit sum equal to its assumed value, namely $16.$ Therefore the original number is $358.$

Small Hint:

Substitute $x=y-1$ before expanding

Big Hint:

Find only the coefficient of $y^2$ in each power and use the hockey-stick identity

Solution:

The polynomial is $\sum_{k=0}^{17}(-x)^k.$ Since $x=y-1,$ this becomes $$\sum_{k=0}^{17}(1-y)^k.$$ For $k\geq2,$ the coefficient of $y^2$ in $(1-y)^k$ is $\binom{k}{2}.$ Hence $$a_2=\sum_{k=2}^{17}\binom{k}{2} =\binom{18}{3}=816.$$

Small Hint:

If $S$ had six elements, compare the variance of its $64$ subset sums with that of $64$ consecutive integers

Big Hint:

After bounding the size of $S,$ inspect the five-element subsets whose sums exceed the candidate

Solution:

First, $S$ has at most five elements. If it had six elements $s_1,\ldots,s_6,$ then its $64$ subset sums would all be distinct: equality between two subset sums, after cancelling their common elements, would violate the given condition.

Choose a subset uniformly at random and let $X$ be its sum. Then $$\begin{gathered} \operatorname{Var}(X) =\frac14\sum_{i=1}^6s_i^2\\ {}\leq\frac14(10^2+11^2+\cdots+15^2)\\ {}=\frac{955}{4}. \end{gathered}$$ On the other hand, $X$ is uniform on $64$ distinct integers. The least possible variance for $64$ distinct integers occurs when they are consecutive, and is $$\frac{64^2-1}{12}=\frac{1365}{4},$$ a contradiction.

A set with at most four elements has sum at most $12+13+14+15=54.$ There are only seven five-element subsets of $\{1,\ldots,15\}$ whose sums are at least $62.$ Each fails, as witnessed by the following equal sums:

$\{8,12,13,14,15\}:$ $13+14=12+15.$ $\{9,11,13,14,15\}:$ $11+13=9+15.$ $\{9,12,13,14,15\}:$ $13+14=12+15.$

$\{10,11,12,14,15\}:$ $11+14=10+15.$ $\{10,11,13,14,15\}:$ $11+13=10+14.$

$\{10,12,13,14,15\}:$ $12+13=10+15.$ $\{11,12,13,14,15\}:$ $12+13=11+14.$

Thus the answer is at most $61.$

The set $\{8,11,13,14,15\}$ attains $61.$ Its $32$ subset sums, in order, are $$\begin{gathered} 0,8,11,13,14,15,19,21,\\ 22,23,24,25,26,27,28,29,\\ 32,33,34,35,36,37,38,39,\\ 40,42,46,47,48,50,53,61, \end{gathered}$$ all distinct. Equal sums from arbitrary subsets would, after deleting their intersection, give equal sums from disjoint subsets, so this verifies the required property.

Small Hint:

Compare the numbers of $\mathrm{HT}$ and $\mathrm{TH}$ transitions to determine the first and last tosses

Big Hint:

Translate the $\mathrm{HH}$ and $\mathrm{TT}$ counts into totals distributed among alternating runs

Solution:

Since there are four $\mathrm{TH}$ transitions and three $\mathrm{HT}$ transitions, every valid sequence starts with $\mathrm{T}$ and ends with $\mathrm{H}.$ It therefore has four $\mathrm{T}$-runs and four $\mathrm{H}$-runs, alternating.

If the $\mathrm{H}$-runs have total length $h,$ then the number of $\mathrm{HH}$ transitions is $h-4.$ Thus $h=6,$ and the positive lengths of the four $\mathrm{H}$-runs can be chosen in $\binom{6-1}{4-1}=\binom53=10$ ways. Similarly, five $\mathrm{TT}$ transitions mean that the four $\mathrm{T}$-runs have total length $9,$ giving $\binom{9-1}{4-1}=\binom83=56$ choices. The alternating order is fixed, so the number of sequences is $10\cdot56=560.$

Small Hint:

Let the side lengths be $a,$ $b,$ and $c$ and use a vector formula for the distance between skew lines

Big Hint:

Taking reciprocals of the squared distances makes the equations linear in $\frac{1}{a^2},$ $\frac{1}{b^2},$ and $\frac{1}{c^2}$

Solution:

Let the side lengths be $a,$ $b,$ and $c.$ For example, the distance from the space diagonal with direction $(a,b,c)$ to a nonintersecting edge parallel to the $a$-direction is $$\frac{bc}{\sqrt{b^2+c^2}}.$$ The other two distances are obtained cyclically. We may assign the three given distances to these three directions in the listed order, since permuting them only permutes the side lengths.

Put $X=\frac{1}{a^2},$ $Y=\frac{1}{b^2},$ and $Z=\frac{1}{c^2}.$ Taking reciprocal squares gives $$\begin{aligned} Y+Z&=\frac1{20},\\ X+Z&=\frac{13}{900},\\ X+Y&=\frac2{45}. \end{aligned}$$ Solving, $$\begin{aligned} X&=\frac1{225},\\ Y&=\frac1{25},\\ Z&=\frac1{100}. \end{aligned}$$ Hence the side lengths are $15,$ $5,$ $10,$ and the volume is $15\cdot5\cdot10=750.$

Small Hint:

The intersection of the two median lines is the centroid

Big Hint:

Parametrize $A$ and $B$ from the centroid along direction vectors $(1,1)$ and $(1,2)$

Solution:

The median lines meet at the centroid $G=(-1,2).$ For real $u$ and $v,$ write $$\begin{aligned} A&=G+(u,u),\\ B&=G+(v,2v). \end{aligned}$$ Since $A+B+C=3G,$ $$C=G-(u,u)-(v,2v).$$ The condition $AC\perp BC$ gives $$\begin{aligned} &(2u+v,2u+2v)\\ &\qquad\mathbin{\cdot}(u+2v,u+4v)=0, \end{aligned}$$ or $$4u^2+15uv+10v^2=0.$$ Meanwhile $AB=60$ gives $$\begin{aligned} &(u-v)^2+(u-2v)^2\\ &\qquad=2u^2-6uv+5v^2\\ &\qquad=3600. \end{aligned}$$ Subtracting half of the first equation from the second yields $-\frac{27}{2}uv=3600,$ so $uv=-\frac{800}{3}.$

Using the two perpendicular legs from $C,$ the area is half the absolute determinant: $$\begin{aligned} [ABC] &=\frac12\left| \det\begin{pmatrix}2u+v&2u+2v\\u+2v&u+4v\end{pmatrix} \right|\\ &=\frac12|3uv| =400. \end{aligned}$$