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

Numbers that are both squares and cubes are sixth powers

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 $528.$

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

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

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 $$\begin{aligned}6(43)(3)+2(27)&=774+54\\&=828.\end{aligned}$$

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

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

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 $117.$

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

Combine the first two fractions before clearing denominators

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 $13.$

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

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

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 $$\frac n{75}=\frac{2^4\cdot3^4\cdot5^2}{3\cdot5^2}=2^4\cdot3^3=432.$$

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

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

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 $\frac{630}{0.75}=840.$

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

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

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 $a+c=11+78=89.$

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

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

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 $$\frac{8!}{3!\,3!\,2!}=560.$$

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

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

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 $i+j=9+64=73.$

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

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

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 $\operatorname{lcm}(18,48)=144.$

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

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

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 $23.$

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

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

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 $a+b+c+d=720.$