Small Hint:

Count squares and cubes up to a candidate endpoint by inclusion-exclusion

Big Hint:

Numbers that are both squares and cubes are sixth powers

Solution:

Through $529,$ there are $\lfloor\sqrt{529}\rfloor=23$ squares, $8$ cubes, and $2$ sixth powers counted in both groups. Thus the number of allowed terms at most $529$ is $$529-23-8+2=500.$$ But $529=23^2$ is excluded, while $528$ is neither a square nor a cube. Therefore the $500$th term is $528.$

Small Hint:

Express $52\pm6\sqrt{43}$ as squares of conjugate radical expressions

Big Hint:

After taking the $\frac{3}{2}$ powers, expand the difference of the two cubes symmetrically

Solution:

Since $$52\pm6\sqrt{43}=(\sqrt{43}\pm3)^2$$ and $\sqrt{43}\gt3,$ the expression is $(\sqrt{43}+3)^3-(\sqrt{43}-3)^3.$ Using $(x+y)^3-(x-y)^3=6x^2y+2y^3$ with $x=\sqrt{43}$ and $y=3$ gives $$\begin{aligned}6(43)(3)+2(27)&=774+54\\&=828.\end{aligned}$$

Small Hint:

Use $\frac{180(n-2)}{n}$ for the interior angle of a regular $n$-gon

Big Hint:

After simplifying, set $t=118-s$ and use the positivity of $t$

Solution:

The angle condition gives $$\frac{\frac{r-2}{r}}{\frac{s-2}{s}}=\frac{59}{58},$$ which simplifies to $r(118-s)=116s.$ Put $t=118-s.$ Then $$r=\frac{13688}{t}-116.$$ In particular, $t$ is a positive divisor of $13688,$ and maximizing $s=118-t$ means taking the least possible $t.$ The choice $t=1$ gives $s=117$ and the positive integer $r=13572,$ which also satisfies $r\geq s.$ Hence the largest possible $s$ is $117.$

Small Hint:

Set $y=x^2-10x$ so the three denominators differ only by constants

Big Hint:

Combine the first two fractions before clearing denominators

Solution:

Set $y=x^2-10x.$ Combining the first two fractions and clearing the nonzero denominators gives an equality between $(y-37)(y-69)$ and $(y-29)(y-45).$ Expanding and canceling $y^2$ yields $y=39.$ Thus $x^2-10x-39=0,$ so $(x-13)(x+3)=0.$ The positive solution is $13.$

Small Hint:

Factor $75$ into possible products of numbers one greater than prime exponents

Big Hint:

The exponent pattern $(4,4,2)$ can include the required factors $3$ and $5^2$ while assigning the largest exponents to the smallest primes

Solution:

The multiplicative partitions of $75$ give exponent patterns $(74),$ $(24,2),$ $(14,4),$ and $(4,4,2).$ A number divisible by $75=3\cdot5^2$ needs both primes $3$ and $5,$ so the one-prime pattern is impossible. The smallest candidates from the two-prime patterns are $3^{24}5^2$ and $3^{14}5^4,$ respectively. The smallest three-prime candidate is $$n=2^4\cdot3^4\cdot5^2.$$ It has $(4+1)(4+1)(2+1)=75$ divisors, and each two-prime candidate is larger because $$\frac{3^{24}5^2}{n}=\frac{3^{20}}{16}\gt1$$ and $$\frac{3^{14}5^4}{n}=\frac{3^{10}5^2}{16}\gt1.$$ Therefore $$\frac n{75}=\frac{2^4\cdot3^4\cdot5^2}{3\cdot5^2}=2^4\cdot3^3=432.$$

Small Hint:

First determine how many of the original tagged fish remain in September

Big Hint:

Use the sample’s tagged fraction to estimate the September population, then identify the $60\%$ that were present in May

Solution:

Of the $60$ tagged fish, $75\%$ remain in September, so $45$ tagged fish remain. The sample estimates that tagged fish form $\frac{3}{70}$ of the September population, making that population $$45\left(\frac{70}{3}\right)=1050.$$ Of those fish, $60\%$ were present in May, so $630$ surviving May fish remain. These are $75\%$ of the May population. Thus the estimated May population is $\frac{630}{0.75}=840.$

Small Hint:

Find the unit vectors from $P$ toward $Q$ and $R$

Big Hint:

The internal angle-bisector direction is the sum of the two unit vectors

Solution:

We have $\overrightarrow{PQ}=(-7,-24)$ with length $25,$ and $\overrightarrow{PR}=(9,-12)$ with length $15.$ The sum of their unit vectors is $$\begin{aligned}v&=\left(-\frac7{25},-\frac{24}{25}\right)\\&\quad+\left(\frac35,-\frac45\right)\\&=\frac1{25}(8,-44),\end{aligned}$$ so the angle bisector has direction $(2,-11).$ A normal vector is $(11,2).$ Through $P=(-8,5),$ its equation is $$11(x+8)+2(y-5)=0,$$ or $11x+2y+78=0.$ Hence $a+c=11+78=89.$

Small Hint:

Within each column, the bottom-to-top order is forced

Big Hint:

Encode an order only by the sequence of chosen columns, with multiplicities $3,$ $3,$ and $2$

Solution:

Once the chosen column is known at each shot, the target within that column is forced. Thus every valid order corresponds to an arrangement of three symbols from the first column, three from the second, and two from the third. The number of such arrangements is $$\frac{8!}{3!\,3!\,2!}=560.$$

Small Hint:

Count valid length-$n$ toss strings according to whether they end in tails or heads

Big Hint:

The resulting recurrence is Fibonacci-like, with initial counts $2$ and $3$

Solution:

Let $u_n$ be the number of length-$n$ toss strings with no consecutive heads. A valid string ending in tails is obtained by appending $\mathrm{T}$ to any valid length-$(n-1)$ string, while one ending in heads is obtained by appending $\mathrm{TH}$ to any valid length-$(n-2)$ string. Hence $u_n=u_{n-1}+u_{n-2},$ with $u_1=2$ and $u_2=3.$ This gives $u_{10}=144.$ The probability is $\frac{144}{2^{10}}=\frac{9}{64},$ so $i+j=9+64=73.$

Small Hint:

Write the roots as exponentials whose arguments are multiples of $\frac{2\pi}{18}$ and $\frac{2\pi}{48}$

Big Hint:

The sums of those arguments generate all multiples of $\frac{2\pi}{\operatorname{lcm}(18,48)}$

Solution:

The arguments of products in $C$ are $$2\pi\left(\frac a{18}+\frac b{48}\right)=\frac{2\pi(8a+3b)}{144}.$$ Since $\gcd(8,3)=1,$ the residues $8a+3b$ generate every residue modulo $144.$ Thus $C$ is precisely the set of $144$th roots of unity. Equivalently, its order is $\operatorname{lcm}(18,48)=144.$

Small Hint:

Compare $n!$ with products of $n-3$ consecutive integers beginning at $4$ and at $5$

Big Hint:

The product beginning at $5$ equals $\frac{(n+1)!}{4!}$

Solution:

The $n-3$ consecutive integers beginning at $4$ have product $$4\cdot5\cdots n=\frac{n!}{6},$$ while those beginning at $5$ have product $$\begin{aligned}5\cdot6\cdots(n+1)&=\frac{(n+1)!}{24}\\&=\frac{n+1}{24}n!.\end{aligned}$$ For $n=23,$ the latter product equals $n!,$ so $23$ works. For every $n\geq24,$ the product beginning at $4$ is below $n!,$ the product beginning at $5$ is above $n!,$ and the product strictly increases with its initial term. Hence no $n\geq24$ works, and the largest possible value is $23.$

Small Hint:

Group the chords by the number of vertex steps $k,$ where $k$ is $1,$ $2,$ $\ldots,$ or $6$

Big Hint:

For $k\lt6$ there are $12$ chords of length $24\sin(\frac{k\pi}{12}),$ while there are $6$ diameters

Solution:

For $k$ equal to $1,$ $2,$ $\ldots,$ and $5,$ there are $12$ chords of length $24\sin(\frac{k\pi}{12}),$ and there are $6$ diameters of length $24.$ The five chord lengths are $$\begin{gathered}6(\sqrt6-\sqrt2),\quad12,\quad12\sqrt2,\\12\sqrt3,\quad6(\sqrt6+\sqrt2).\end{gathered}$$ Their sum $U$ is $$\begin{aligned}U&=12+12\sqrt2\\&\quad+12\sqrt3+12\sqrt6.\end{aligned}$$ Therefore the total is $$\begin{aligned}12U+6(24)&=288+144\sqrt2\\&\quad+144\sqrt3\\&\quad+144\sqrt6.\end{aligned}$$ Hence $a=288$ and $b=c=d=144,$ so $a+b+c+d=720.$

Small Hint:

Compare the number of digits of $9^k$ with that of $9^{k-1}$

Big Hint:

A multiplication by $9$ produces a leading $9$ exactly when the digit count does not increase

Solution:

For $k\geq1,$ the number $9^k$ begins with $9$ exactly when it has the same number of digits as $9^{k-1}.$ Indeed, if $9^{k-1}$ has $d$ digits and multiplication by $9$ creates no new digit, then $9^k\geq9\cdot10^{d-1},$ so its leading digit is $9.$ If multiplication does create a new digit, then $9^k\lt9\cdot10^d,$ so its leading digit is at most $8.$

Starting from the one-digit number $9^0,$ the digit count reaches $3817$ after $4000$ multiplications. Thus it increases on $3816$ steps and stays unchanged on $4000-3816=184$ steps. Since $9^0=1$ does not begin with $9,$ exactly $184$ elements of $T$ do.

Small Hint:

After $AP$ and $BP$ are joined, vertices $A$ and $B$ become one vertex; determine all six edge lengths of the tetrahedron

Big Hint:

Place $C,$ $D,$ and the joined vertex in one coordinate plane, then locate $P$ from its equal distances to the other vertices

Solution:

After folding, $A$ and $B$ become one vertex $X.$ Each half-diagonal of the rectangle has length $\frac{\sqrt{939}}{2}.$ Thus $$XP=CP=DP=\frac{\sqrt{939}}2,$$ while $XC=XD=13\sqrt3$ and $CD=12\sqrt3.$

Place $$\begin{aligned}C&=(-6\sqrt3,0,0),\\D&=(6\sqrt3,0,0),\end{aligned}$$ and $X=(0,\sqrt{399},0).$ These coordinates give the required lengths from $X$ to $C$ and $D.$ Because $P$ is equidistant from $C$ and $D,$ write $P=(0,u,h).$ Equating $PC^2$ and $PX^2$ gives $$u=\frac{291}{2\sqrt{399}}.$$ Then $PC^2=\frac{939}{4}$ yields $$h^2=\frac{507}{4}-u^2=\frac{9801}{133},$$ so $h=\frac{99}{\sqrt{133}}.$

The base triangle $XCD$ has area $$\frac12(12\sqrt3)(\sqrt{399})=18\sqrt{133}.$$ Therefore the pyramid’s volume is $$\frac13(18\sqrt{133})\left(\frac{99}{\sqrt{133}}\right)=594.$$

Small Hint:

Let $S_k=ax^k+by^k$ and derive a recurrence using $x+y$ and $xy$

Big Hint:

Use $S_3$ and $S_4$ to solve for the two recurrence coefficients before computing $S_5$

Solution:

Let $S_k=ax^k+by^k,$ $p=x+y,$ and $q=xy.$ Since $x$ and $y$ each satisfy $t^2=pt-q,$ $$S_{k+2}=pS_{k+1}-qS_k.$$ Using $S_1=3,$ $S_2=7,$ $S_3=16,$ and $S_4=42$ gives $$\begin{aligned}7p-3q&=16,\\16p-7q&=42.\end{aligned}$$ Solving yields $p=-14$ and $q=-38.$ Therefore $$\begin{aligned}S_5&=pS_4-qS_3\\&=-14(42)+38(16)\\&=20.\end{aligned}$$