Small Hint:

There are $1012$ people to the left of this person and $1009$ to the right

Big Hint:

Add the two counts, but do not count this one person twice

Solution:

There are $1012$ people to the left of this spot and $1009$ to the right. Add those two groups plus the person themselves: $1012 + 1009 + 1 = 2022.$ Or, just as fast, the two positions overlap on one person, so $1013 + 1010 - 1 = 2022.$ Thus, B is the correct answer.

Small Hint:

Write $10! = 10 \cdot 9 \cdot 8 \cdot 7!$

Big Hint:

Notice that $10 \cdot 9 \cdot 8$ equals $6!$

Solution:

Write $10! = 10 \cdot 9 \cdot 8 \cdot 7! = 720 \cdot 7!.$ But $720 = 6!$ too, so $10! = 6! \cdot 7! = 7! \cdot 6!.$ The two terms are the same. That makes $10! - 7! \cdot 6! = 0.$ Therefore, the answer is B.

Small Hint:

Divide by $2$: the inequality is $|x| \le 3.5\pi$

Big Hint:

$3.5\pi \approx 10.99,$ so $x$ ranges over the integers with $|x| \le 10$

Solution:

Divide by $2$ to get $|x| \le 3.5\pi \approx 10.996.$ The integers that fit run from $-10$ up to $10,$ and there are $21$ of them. Thus, E is the correct answer.

Small Hint:

Group $g$ has $g$ balls, so groups $1$ through $g$ use $\tfrac{g(g+1)}{2}$ balls

Big Hint:

Find the group containing ball $2024,$ then read its bin from $g \bmod 5,$ with group $1$ corresponding to $A$

Solution:

Group $g$ holds $g$ balls, so the first $g$ groups swallow $\tfrac{g(g+1)}{2}$ of them. Now $\tfrac{63 \cdot 64}{2} = 2016$ and $\tfrac{64 \cdot 65}{2} = 2080,$ which puts ball $2024$ in group $64$ (balls $2017$ through $2080$). The bins cycle $A, B, C, D, E,$ so group $g$ lands in bin number $(g - 1) \bmod 5.$ For $g = 64$ that’s $63 \bmod 5 = 3,$ bin $D.$ Therefore, the answer is D.

Small Hint:

The full sum is $1 + 3 + \cdots + 99 = 50^2 = 2500;$ flipping a term $t$ lowers the sum by $2t$

Big Hint:

To go below $0$ with the fewest flips, flip the largest terms; their doubled total must exceed $2500$

Solution:

The full sum is $1 + 3 + \cdots + 99 = 50^2 = 2500.$ Flipping a term $t$ drops the total by $2t,$ so to go negative the flipped terms have to add up to more than $1250.$ The greedy move is to flip the biggest odd numbers: flipping the top $k$ gives $99 + 97 + \cdots = k(100 - k).$ We want $k(100 - k) \gt 1250.$ At $k = 14$ it’s only $1204,$ but at $k = 15$ it jumps to $1275.$ So $15$ flips do it. Thus, B is the correct answer.

Small Hint:

For a fixed area, the perimeter is smallest when the two sides are as close to equal as possible

Big Hint:

$2024 = 2^3 \cdot 11 \cdot 23;$ look for a factor pair near $\sqrt{2024} \approx 45$

Solution:

The perimeter $2(\ell + w)$ with $\ell w = 2024$ is smallest when $\ell$ and $w$ are as close together as possible. Factor $2024 = 2^3 \cdot 11 \cdot 23.$ The divisor pair nearest $\sqrt{2024} \approx 45$ is $44 \times 46,$ which gives perimeter $2(44 + 46) = 180.$ Therefore, the answer is B.

Small Hint:

Factor out $7^{2024}$

Big Hint:

The remaining factor is $1 + 7 + 49;$ check its relationship to $19$

Solution:

Pull out the common power: $7^{2024} + 7^{2025} + 7^{2026}$ $= 7^{2024}(1 + 7 + 49)$ $= 7^{2024} \cdot 57.$ And $57 = 3 \cdot 19,$ so the product is a multiple of $19.$ The remainder is $0.$ Thus, A is the correct answer.

Small Hint:

A number with $d$ divisors has divisor product equal to itself raised to the $\tfrac{d}{2}$ power

Big Hint:

$42 = 2 \cdot 3 \cdot 7$ has $8$ divisors, so $N = 42^4;$ track only the units digit

Solution:

Since $42 = 2 \cdot 3 \cdot 7$ has $(1+1)^3 = 8$ divisors, we can pair each divisor with its complement, so $N = 42^{\frac{8}{2}} = 42^4.$ Only the units digit matters, and $2^4 = 16$ ends in $6,$ so $42^4$ does too. Therefore, the answer is D.

Small Hint:

Mean $0$ means $a + b + c = 0;$ mean $10$ means $a^2 + b^2 + c^2 = 30$

Big Hint:

Use $(a + b + c)^2$ $= a^2 + b^2 + c^2$ $+ 2(ab + bc + ca)$

Solution:

The means tell us $a + b + c = 0$ and $a^2 + b^2 + c^2 = 30.$ Square the first: $(a+b+c)^2$ $= a^2 + b^2 + c^2$ $+ 2(ab + bc + ca),$ so $0 = 30 + 2(ab + bc + ca)$ and $ab + bc + ca = -15.$ Their mean is $-\frac{15}{3} = -5.$ Thus, A is the correct answer.

Small Hint:

Area ratios are unchanged by an affine map, so compute with $A = (0,0),$ $B = (1,0),$ $C = (1,1),$ $D = (0,1)$

Big Hint:

Find $F$ as the intersection of line $AC$ with line $EB,$ then use the shoelace formula

Solution:

Area ratios don’t change under an affine map, so drop in convenient coordinates: $A = (0,0),$ $B = (1,0),$ $C = (1,1),$ $D = (0,1),$ which makes $E = (0, \tfrac12).$ Line $AC$ is $y = x,$ and line $EB$ runs from $(0, \tfrac12)$ to $(1, 0);$ they cross at $F = (\tfrac13, \tfrac13).$ The shoelace formula gives quadrilateral $CDEF$ area $\tfrac{5}{12}$ and triangle $CFB$ area $\tfrac13.$ So the ratio is $\tfrac{5}{12} : \tfrac13 = 5 : 4.$ Therefore, the answer is A.

Small Hint:

Set $X = (0,0),$ $Y = (8,0),$ $W = (0,4),$ $Z = (8,4),$ with $M = (m, 0)$ and $A = (8, a)$

Big Hint:

Perpendicularity gives $m(8 - m) = 4a;$ equal areas give $2m = 4(4 - a)$

Solution:

Set $X = (0,0),$ $Y = (8,0),$ $W = (0,4),$ $Z = (8,4),$ so $M = (m, 0)$ and $A = (8, a).$ The right angle means $\overrightarrow{MW} \cdot \overrightarrow{MA} = 0,$ which gives $-m(8 - m) + 4a = 0,$ that is $m(8 - m) = 4a.$ Equal areas $[WXM] = 2m$ and $[WAZ] = 4(4 - a)$ force $m = 8 - 2a,$ so $a = \tfrac{8 - m}{2}.$ Substitute back and $(8 - m)(2 - m) = 0.$ Taking $M \ne Y$ leaves $m = 2$ and $a = 3.$ Then $[WMA] = 32$ $- [WXM]$ $- [MYA]$ $- [AZW]$ $= 32 - 4 - 9 - 4$ $= 15.$ Thus, C is the correct answer.

Small Hint:

Give each student the set of languages they speak; the condition means no set contains another

Big Hint:

Equal-size distinct sets never contain one another, so you need $\binom{n}{k} \ge 100$ for some $k$

Solution:

Give each student the set of languages they speak. The condition says no one’s set sits inside another’s. Everyone speaks the same number $k$ of languages, and two distinct $k$-element sets can never contain each other, so all we need is $100$ different $k$-subsets of the $n$ languages, i.e. $\binom{n}{k} \ge 100.$ With $n = 8$ the best we can manage is $\binom{8}{4} = 70,$ short of $100.$ But $\binom{9}{4} = 126 \ge 100.$ So $9$ languages are both enough and necessary. Therefore, the answer is A.

Small Hint:

$1183 = 7 \cdot 13^2,$ so $\sqrt{1183} = 13\sqrt{7}$

Big Hint:

Then $\sqrt{x} = a\sqrt{7}$ and $\sqrt{y} = b\sqrt{7}$ with $a + b = 13;$ minimize $x + y = 7(a^2 + b^2)$

Solution:

Since $1183 = 7 \cdot 13^2,$ we have $\sqrt{1183} = 13\sqrt7.$ Squaring the given equation shows that $\sqrt{xy}$ is rational. Thus $x$ and $y$ have the same squarefree part: write $x=da^2$ and $y=db^2,$ where $d$ is squarefree and $a,b$ are positive integers. Then $(a+b)\sqrt d=13\sqrt7,$ so $d=7$ and $a+b=13.$ Therefore $x+y=7(a^2+b^2),$ which is smallest when $a$ and $b$ are as close as possible. Taking $a=6$ and $b=7$ gives $x+y=7(36+49)=595.$ Thus, B is the correct answer.

Small Hint:

$B$ is a square of diagonal $16,$ and $T$ is the annulus $18 \le x^2 + y^2 \le 32$

Big Hint:

The outer radius $\sqrt{32} = 4\sqrt2$ equals the distance from the origin to each edge of $B,$ so $T$ sits entirely inside $B$

Solution:

$B$ is the square $|x| + |y| \le 8,$ with area $2 \cdot 8^2 = 128.$ The target condition $(x^2 + y^2 - 25)^2 \le 49$ unpacks to $|x^2 + y^2 - 25| \le 7,$ that is $18 \le x^2 + y^2 \le 32,$ an annulus of area $\pi(32 - 18) = 14\pi.$ Does it fit inside $B?$ The distance from the origin to an edge $x + y = 8$ is $\tfrac{8}{\sqrt2} = 4\sqrt2 = \sqrt{32},$ exactly the outer radius, so yes, the annulus sits inside the square. The probability is $\tfrac{14\pi}{128} = \tfrac{7}{64}\pi,$ giving $m + n = 71.$ Therefore, the answer is B.

Small Hint:

The six given numbers sum to $24.8,$ so $x + y + z$ must make the total a multiple of $9$

Big Hint:

Range $7$ fixes the smallest and largest values; then require the $5$th smallest number to be a positive integer

Solution:

The six fixed numbers total $24.8.$ If the mean is the integer $k,$ then $x+y+z=9k-24.8.$ Because the fixed entries already run from $1$ to $7,$ the range condition gives three cases.

If $z\le7,$ then $x=0.$ The bounds on $y+z$ force $k=3$ or $4.$ For $k=3,$ the median is $2.2;$ for $k=4,$ we have $y+z=11.2,$ so $y\ge4.2$ and the median is integral only when $y=5,$ giving $(x,y,z)=(0,5,6.2).$

If $x\ge1,$ then $z=8.$ Here $k=4$ gives median $3.2.$ For $k=5,$ we have $x+y=12.2;$ the median is integral only when $x=6,$ giving $(6,6.2,8).$

The remaining case has $0<x<1$ and $z=x+7.$ The total lies between $31.8$ and $41.8,$ so $k=4$ and $y=4.2-2x.$ The median can be an integer only when $y=4,$ giving $(0.1,4,7.1).$ Hence exactly $3$ ordered triples work. Thus, C is the correct answer.

Small Hint:

Each move turns $4$ numbers into $1,$ so the count drops by $3;$ maximizing the count means minimizing the moves

Big Hint:

There are $1012$ even entries, and any move decreases their number by at most $3$

Solution:

Among $1, \ldots, 2024$ there are $1012$ even numbers and $1012$ odd numbers. Each operation replaces $4$ entries by $1,$ so the total count falls by $3.$ If a move consumes $e$ even entries, its output is either odd, reducing the even count by $e,$ or even, reducing it by $e-1.$ In either case the even count falls by at most $3.$ Therefore eliminating all $1012$ even entries takes at least $\lceil\frac{1012}{3}\rceil=338$ moves. This is achievable: use $337$ sums containing one odd and three evens, then one sum containing three odds and the final even. Every output is odd. Thus the maximum remaining count is $2024-3\cdot338=1010.$ Therefore, the answer is A.