2024 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

What is the value of $9901\cdot101-99\cdot10101?$

$2$

$20$

$21$

$200$

$2020$

Answer: A

Concepts:whole number operations

Difficulty rating: 870

Small Hint:

$9901\cdot101=9901\cdot100+9901$

Big Hint:

$99\cdot10101=100\cdot10101-10101$

Solution:

Directly, $9901\cdot101=990100$ $+9901=1000001$ and $99\cdot10101=1010100$ $-10101=999999.$ Their difference is $1000001-999999=2.$ Thus, the correct answer is A.

2.

A model used to estimate the time it will take to hike to the top of a mountain on a trail is of the form $T=aL+bG,$ where $a$ and $b$ are constants, $T$ is the time in minutes, $L$ is the length of the trail in miles, and $G$ is the altitude gain in feet. The model estimates that it will take $69$ minutes to hike to the top if a trail is $1.5$ miles long and ascends $800$ feet, as well as if a trail is $1.2$ miles long and ascends $1100$ feet.

How many minutes does the model estimate it will take to hike to the top if the trail is $4.2$ miles long and ascends $4000$ feet?

$240$

$246$

$252$

$258$

$264$

Answer: B

Concepts:system of equations substitution

Difficulty rating: 990

Small Hint:

Set the two estimates equal: $1.5a+800b=1.2a+1100b$

Big Hint:

This gives $a=1000b,$ so $69=2300b$

Solution:

From $1.5a+800b=1.2a+1100b$ we get $0.3a=300b,$ so $a=1000b.$ Then $69=1.5a+800b$ $=1500b+800b=2300b,$ giving $b=0.03$ and $a=30.$ For $L=4.2,\ G=4000,$ $T=30(4.2)$ $+0.03(4000)$ $=126+120=246.$ Thus, the correct answer is B.

3.

The number $2024$ is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?

$20$

$21$

$22$

$23$

$24$

Answer: B

Concepts:optimization bounding to limit cases

Difficulty rating: 1100

Small Hint:

The largest two-digit number is $99,$ so many numbers are needed

Big Hint:

$20\cdot99=1980,$ which falls short; can $21$ numbers reach exactly $2024?$

Solution:

Each number is at most $99,$ so $k$ numbers sum to at most $99k.$ Since $99\cdot20=1980\lt2024,$ at least $21$ numbers are required. With $21,$ we can use twenty $99$ s and one $44$ : $20\cdot99+44=1980+44$ $=2024.$ Thus, the correct answer is B.

4.

What is the least value of $n$ such that $n!$ is a multiple of $2024?$

$11$

$21$

$22$

$23$

$253$

Answer: D

Concepts:prime factorization factorial

Difficulty rating: 1180

Small Hint:

Factor $2024=2^3\cdot11\cdot23$

Big Hint:

$n!$ must contain the prime $23,$ so $n\ge23$

Solution:

Factoring, $2024=2^3\cdot11\cdot23.$ The factorial $n!$ contains the prime $23$ only when $n\ge23.$ At $n=23,$ the product $23!$ already includes $23,\ 11,$ and plenty of factors of $2,$ so $23!$ is a multiple of $2024.$ Thus, the correct answer is D.

5.

A data set containing $20$ numbers, some of which are $6,$ has mean $45.$ When all the $6$ s are removed, the data set has mean $66.$ How many $6$ s were in the original data set?

$4$

$5$

$6$

$7$

$8$

Answer: D

Concepts:mean linear equation

Difficulty rating: 1230

Small Hint:

The $20$ numbers sum to $20\cdot45=900$

Big Hint:

If there are $k$ sixes, the remaining $20-k$ numbers sum to $900-6k=66(20-k)$

Solution:

The full set sums to $20\cdot45=900.$ Removing $k$ sixes leaves $20-k$ numbers summing to $900-6k,$ with mean $66,$ so $900-6k=66(20-k)$ $=1320-66k.$ Then $60k=420,$ giving $k=7.$ Thus, the correct answer is D.

6.

The product of three integers is $60.$ What is the least possible positive sum of the three integers?

$2$

$3$

$5$

$6$

$13$

Answer: B

Concepts:factor optimization casework

Difficulty rating: 1350

Small Hint:

Integers may be negative; the product stays positive with two negative factors

Big Hint:

Writing the numbers as $-p,-q,r$ with $pqr=60,$ minimize $r-p-q$ while keeping it positive

Solution:

To obtain a small positive sum, use two negative integers $-p,-q$ and one positive integer $r,$ where $pqr=60.$ Up to order, the positive factor triples of $60$ are $\begin{gathered} (1,1,60),(1,2,30),\\ (1,3,20),(1,4,15),\\ (1,5,12),(1,6,10),\\ (2,2,15),(2,3,10),\\ (2,5,6),(3,4,5). \end{gathered}$ A positive value of $r-p-q$ is smallest when the largest factor is chosen as $r.$ The positive values from the list are $58,27,16,10,6,3,11,5;$ the last two triples give negative values. Thus the least positive sum is $10-1-6=3.$ (Three positive integers have sum at least $3,$ and three negative integers cannot have positive product.) Thus, the correct answer is B.

7.

In $\triangle ABC,$ $\angle ABC=90^\circ$ and $BA=BC=\sqrt2.$ Points $P_1,$ $P_2,$ $\ldots,$ $P_{2024}$ lie on hypotenuse $AC$ so that $AP_1=P_1P_2$ $=P_2P_3$ $=\cdots$ $=P_{2023}P_{2024}$ $=P_{2024}C.$

What is the length of the vector sum $\begin{aligned} &\vec{BP_1}+\vec{BP_2}+\vec{BP_3} \\ &\quad {}+\cdots+\vec{BP_{2024}}? \end{aligned}$

$1011$

$1012$

$2023$

$2024$

$2025$

Answer: D

Concepts:vector median (geometry)symmetry

Difficulty rating: 1430

Small Hint:

The points are symmetric about the midpoint $M$ of $AC,$ so the vectors pair up

Big Hint:

The sum equals $2024\,\vec{BM};$ the median to the hypotenuse of a right triangle is half the hypotenuse

Solution:

The points $P_k$ are symmetric about the midpoint $M$ of $AC,$ so pairing $P_k$ with its mirror gives $\vec{BP_k}+\vec{BP_{2025-k}}=2\,\vec{BM}.$ Hence the whole sum is $2024\,\vec{BM}.$ In a right triangle the median to the hypotenuse has length half the hypotenuse; here $AC=2,$ so $BM=1.$ The length of the sum is $2024\cdot1=2024.$ Thus, the correct answer is D.

8.

How many angles $\theta$ with $0\le\theta\le2\pi$ satisfy $\log(\sin(3\theta))+\log(\cos(2\theta))=0?$

$0$

$1$

$2$

$3$

$4$

Answer: A

Concepts:logarithm trigonometry bounding to limit cases

Difficulty rating: 1480

Small Hint:

$\log(\sin3\theta)+\log(\cos2\theta)$ $=\log(\sin3\theta\cos2\theta),$ which must equal $0$

Big Hint:

A product of two numbers each at most $1$ equals $1$ only when both equal $1$

Solution:

The equation means $\sin(3\theta)\cos(2\theta)=1$ with both factors positive (for the logs to be defined). Since $\sin(3\theta)\le1$ and $\cos(2\theta)\le1,$ their product is $1$ only if $\sin(3\theta)=1$ and $\cos(2\theta)=1$ simultaneously. But $\cos(2\theta)=1$ forces $\theta\in\{0,\pi,2\pi\},$ where $\sin(3\theta)=0\ne1.$ No angle works. Thus, the correct answer is A.

9.

Let $M$ be the greatest integer such that both $M+1213$ and $M+3773$ are perfect squares. What is the units digit of $M?$

$1$

$2$

$3$

$6$

$8$

Answer: E

Concepts:difference of squares perfect square optimization

Difficulty rating: 1510

Small Hint:

Let $M+1213=a^2$ and $M+3773=b^2;$ subtract to get $b^2-a^2=2560$

Big Hint:

$(b-a)(b+a)=2560$ with both factors even; to maximize $M,$ make $b-a$ as small as possible

Solution:

Write $M+1213=a^2$ and $M+3773=b^2,$ so $b^2-a^2=2560,$ i.e. $(b-a)(b+a)=2560.$ Both factors have the same parity, hence both even. To maximize $a$ (and thus $M$ ), minimize $b-a:$ take $b-a=2,\ b+a=1280,$ so $a=639.$ Then $M=639^2-1213$ $=408321-1213$ $=407108,$ whose units digit is $8.$ Thus, the correct answer is E.

10.

Let $\alpha$ be the radian measure of the smallest angle in a $3\text{-}4\text{-}5$ right triangle. Let $\beta$ be the radian measure of the smallest angle in a $7\text{-}24\text{-}25$ right triangle. In terms of $\alpha,$ what is $\beta?$

$\dfrac{\alpha}{3}$

$\alpha-\dfrac{\pi}{8}$

$\dfrac{\pi}{2}-2\alpha$

$\dfrac{\alpha}{2}$

$\pi-4\alpha$

Answer: C

Concepts:trigonometric identity right triangle

Difficulty rating: 1570

Small Hint:

$\tan\alpha=\dfrac34;$ apply the double-angle formula for $\tan2\alpha$

Big Hint:

$\tan\beta=\dfrac{7}{24}$ is the reciprocal of $\tan2\alpha$

Solution:

The smallest angle of the $3\text{-}4\text{-}5$ triangle has $\tan\alpha=\tfrac34.$ Then $\begin{aligned} &\tan2\alpha=\frac{2\cdot\frac34}{1-\frac{9}{16}} \\ &=\frac{\frac{3}{2}}{\frac{7}{16}}=\frac{24}{7}. \end{aligned}$ The smallest angle of the $7\text{-}24\text{-}25$ triangle has $\tan\beta=\tfrac{7}{24}=\cot2\alpha$ $=\tan\!\left(\tfrac{\pi}{2}-2\alpha\right).$ Hence $\beta=\tfrac{\pi}{2}-2\alpha.$ Thus, the correct answer is C.

11.

There are exactly $K$ positive integers $b$ with $5\le b\le2024$ such that the base- $b$ integer $2024_b$ is divisible by $16$ (where $16$ is in base ten). What is the sum of the digits of $K?$

$16$

$17$

$18$

$20$

$21$

Answer: D

Concepts:number base modular arithmetic counting integers in a range

Difficulty rating: 1630

Small Hint:

$2024_b=2b^3+2b+4,$ so you need $2(b^3+b+2)$ to be divisible by $16$

Big Hint:

Reduce $b^3+b+2\pmod 8$ and count $b$ in each valid residue class of $[5,2024]$

Solution:

Here $2024_b=2b^3+2b+4$ $=2(b^3+b+2),$ so $2024_b$ is divisible by $16$ exactly when $b^3+b+2$ is divisible by $8.$ Checking residues $\bmod 8,$ $b^3+b+2\equiv0$ precisely for $b\equiv3,6,7\pmod8.$

Counting $b$ in $[5,2024]:$ residue $3$ gives $11,\ldots,2019$ ( $252$ values), residue $6$ gives $6,\ldots,2022$ ( $253$ values), and residue $7$ gives $7,\ldots,2023$ ( $253$ values). So $K=252+253+253=758,$ and its digit sum is $7+5+8=20.$

Thus, the correct answer is D.

12.

The first three terms of a geometric sequence are the integers $a,$ $720,$ and $b,$ where $a\lt720\lt b.$ What is the sum of the digits of the least possible value of $b?$

$9$

$12$

$16$

$18$

$21$

Answer: E

Concepts:geometric sequence factor optimization

Difficulty rating: 1630

Small Hint:

The middle term gives $720^2=ab,$ so $ab=518400$

Big Hint:

To minimize $b,$ take the largest divisor $a\lt720$ of $518400$

Solution:

Since the terms are geometric, $720^2=ab,$ so $ab=518400=2^8\cdot3^4\cdot5^2.$ Because $b=\frac{518400}{a},$ minimizing $b$ means finding the smallest divisor of $518400$ greater than $720.$ There is no divisor strictly between $720$ and $768:$ if its exponent of $5$ is $2,1,$ or $0,$ then after dividing by $25,5,$ or $1,$ respectively, a number of the form $2^i3^j$ would have to lie in $(28.8,30.72),\ (144,153.6),$ or $(720,768);$ the allowed powers $i\le8,j\le4$ give none. Since $768=2^8\cdot3$ is a divisor, it is the least possible $b.$ Its paired divisor is $a=\frac{518400}{768}=675,$ and the digit sum of $b$ is $7+6+8=21.$ Thus, the correct answer is E.

13.

The graph of $y=e^{x+1}+e^{-x}-2$ has an axis of symmetry. What is the reflection of the point $\left(-1,\tfrac12\right)$ over this axis?

$\left(-1,-\tfrac32\right)$

$(-1,0)$

$\left(-1,\tfrac12\right)$

$\left(0,\tfrac12\right)$

$\left(3,\tfrac12\right)$

Answer: D

Concepts:function symmetry

Difficulty rating: 1660

Small Hint:

The axis of symmetry is the vertical line $x=c$ where the function attains its minimum

Big Hint:

Set the derivative $e^{x+1}-e^{-x}=0$ to find $c,$ then reflect $x=-1$ across it

Solution:

The curve $y=e^{x+1}+e^{-x}-2$ is symmetric about the vertical line through its minimum. Setting the derivative $e^{x+1}-e^{-x}=0$ gives $x+1=-x,$ so $x=-\tfrac12.$ Reflecting $\left(-1,\tfrac12\right)$ across $x=-\tfrac12$ keeps the $y$ -coordinate and sends $x=-1$ to $x=0.$ The image is $\left(0,\tfrac12\right).$ Thus, the correct answer is D.

14.

The numbers, in order, of each row and the numbers, in order, of each column of a $5\times5$ array of integers form an arithmetic progression of length $5.$ The numbers in positions $(5,5),$ $(2,4),$ $(4,3),$ and $(3,1)$ are $0,$ $48,$ $16,$ and $12,$ respectively. What number is in position $(1,2)?$

$\begin{bmatrix} \cdot & ? & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & 48 & \cdot \\ 12 & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & 16 & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & 0 \end{bmatrix}$

$19$

$24$

$29$

$34$

$39$

Answer: C

Concepts:arithmetic sequence system of equations

Difficulty rating: 1750

Small Hint:

If every row and every column is arithmetic, the entry has the form $\alpha+\beta i+\gamma j+\delta ij$

Big Hint:

Substitute the four known entries into that form and solve for $\alpha,\beta,\gamma,\delta$

Solution:

A grid whose rows and columns are all arithmetic has entries of the bilinear form $a(i,j)=\alpha+\beta i+\gamma j+\delta ij.$ The four givens yield $\alpha+5\beta+5\gamma+25\delta=0,$ $\alpha+2\beta+4\gamma+8\delta=48,$ $\alpha+4\beta+3\gamma+12\delta=16,$ $\alpha+3\beta+\gamma+3\delta=12.$

Solving gives $\delta=-5,\ \beta=5,\$ $\gamma=22,\ \alpha=-10.$ Then $a(1,2)=\alpha+\beta+2\gamma+2\delta$ $=-10+5+44-10$ $=29.$

Thus, the correct answer is C.

15.

The roots of $x^3+2x^2-x+3$ are $p$ , $q$ , and $r$ . What is the value of $(p^2+4)(q^2+4)(r^2+4)?$

$64$

$75$

$100$

$125$

$144$

Answer: D

Concepts:complex number polynomial

Difficulty rating: 1710

Small Hint:

$p^2+4=(p-2i)(p+2i),$ so the product relates to $P(2i)$ and $P(-2i),$ where $P(x)=x^3+2x^2-x+3$

Big Hint:

$P(2i)$ and $P(-2i)$ are complex conjugates; multiply them to get a real number

Solution:

Since $P(x)=(x-p)(x-q)(x-r),$ grouping $p^2+4=(p-2i)(p+2i)$ over all roots gives $\prod(p^2+4)=P(2i)\,P(-2i).$ Compute $P(2i)=-8i-8-2i+3$ $=-5-10i$ and $P(-2i)=8i-8+2i+3$ $=-5+10i.$ Their product is $(-5)^2+10^2=25+100=125.$ Thus, the correct answer is D.

16.

A set of $12$ tokens — $3$ red, $2$ white, $1$ blue, and $6$ black — is to be distributed at random to $3$ game players, $4$ tokens per player. The probability that some player gets all the red tokens, another gets all the white tokens, and the remaining player gets the blue token can be written as $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n?$

$387$

$388$

$389$

$390$

$391$

Answer: C

Concepts:basic probability multiset permutations

Difficulty rating: 1820

Small Hint:

Assign the three roles (all reds, all whites, the lone blue) to the players in $3!$ ways

Big Hint:

The black tokens must then split as $1,2,3$ among the players; divide by $\binom{12}{4,4,4}$

Solution:

Treat all tokens as distinct; the total number of ways to deal $4$ to each player is $\binom{12}{4,4,4}=34650.$ For the favorable event, choose which player gets the reds, whites, and blue in $3!=6$ ways. The red player needs $1$ more token, the white player $2$ more, and the blue player $3$ more, all black; the $6$ black tokens split as $1,2,3$ in $\tfrac{6!}{1!\,2!\,3!}=60$ ways. So the probability is $\dfrac{6\cdot60}{34650}=\dfrac{360}{34650}=\dfrac{4}{385}.$ Then $m+n=4+385=389.$ Thus, the correct answer is C.

17.

Integers $a,$ $b,$ and $c$ satisfy $ab+c=100,$ $bc+a=87,$ and $ca+b=60.$ What is $ab+bc+ca?$

$212$

$247$

$258$

$276$

$284$

Answer: D

Concepts:system of equations factoring casework

Difficulty rating: 1890

Small Hint:

Subtract equations in pairs, e.g. $(ab+c)-(bc+a)$ factors as $(a-c)(b-1)=13$

Big Hint:

$13$ is prime, so $(a-c,\ b-1)$ has few options; test them for integer solutions

Solution:

Subtracting the second equation from the first gives $(a-c)(b-1)=13.$ Hence $\begin{gathered} (a-c,b-1)\in\{(1,13),(13,1),\\ (-1,-13),(-13,-1)\}. \end{gathered}$ The corresponding values of $b$ are $14,2,-12,0.$ Substituting $a=c+(a-c)$ into $ab+c=100$ eliminates the first two cases because they would require $15c=86$ or $3c=74.$ The case $b=-12$ gives $c=-8$ and $a=-9,$ which satisfies all three equations. The case $b=0$ gives $(a,c)=(87,100),$ which fails $ca+b=60.$ Thus the unique integer solution is $(-9,-12,-8),$ and $ab+bc+ca$ $=108+96+72=276.$ Thus, the correct answer is D.

18.

On top of a rectangular card with sides of length $1$ and $2+\sqrt3,$ an identical card is placed so that two of their diagonals line up, as shown ( $AC,$ in this case).

Two congruent rectangular cards sharing the diagonal AC, with the second card rotated.

Continue the process, adding a third card to the second, and so on, lining up successive diagonals after rotating clockwise. In total, how many cards must be used until a vertex of a new card lands exactly on the vertex labeled $B$ in the figure?

$6$

$8$

$10$

$12$

No new vertex will land on $B.$

Answer: A

Concepts:transformation trigonometry

Difficulty rating: 2010

Small Hint:

The diagonal makes angle $\arctan\dfrac{1}{2+\sqrt3}=15^\circ$ with the long side, since $\dfrac{1}{2+\sqrt3}=2-\sqrt3$

Big Hint:

The two diagonals meet at $30^\circ;$ track the unused diagonal as five new cards are added

Solution:

A diagonal makes an angle $\theta$ with a long side, where $\tan\theta=\dfrac{1}{2+\sqrt3}$ $=2-\sqrt3=\tan15^\circ.$ Thus the acute angle between the two diagonals of a card is $2\theta=30^\circ.$

Each new card shares one diagonal with the previous card, and its other diagonal is the next line obtained by turning $30^\circ$ clockwise. All these equal diagonals have the same midpoint and are diameters of one common circle. The line through the original card’s other diagonal, which contains $B,$ is $30^\circ$ counterclockwise from $AC.$ As an unoriented line, this is the same as $150^\circ$ clockwise from $AC.$ Five additions advance the unused diagonal by $5\cdot30^\circ=150^\circ,$ so the sixth card is the first new card with a vertex at $B.$

Thus, the correct answer is A.

19.

Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$ and $DA=5$ with $\angle CDA=120^\circ.$ What is the length of the shorter diagonal of $ABCD?$

$\dfrac{31}{7}$

$\dfrac{33}{7}$

$5$

$\dfrac{39}{7}$

$\dfrac{41}{7}$

Answer: D

Concepts:cyclic quadrilateral law of cosines Ptolemy’s Theorem

Difficulty rating: 1930

Small Hint:

In $\triangle ACD,$ $AC^2=3^2+5^2$ $-2\cdot3\cdot5\cos120^\circ$

Big Hint:

Find $AB$ using $\angle ABC=60^\circ,$ then apply Ptolemy: $AC\cdot BD=AB\cdot CD+BC\cdot DA$

Solution:

In $\triangle ACD,$ the law of cosines gives $AC^2=9+25-2(15)\cos120^\circ$ $=34+15=49,$ so $AC=7.$

Since $ABCD$ is cyclic, $\angle ABC=180^\circ-120^\circ=60^\circ.$ In $\triangle ABC$ with $BC=3$ and $AC=7,$ the law of cosines gives $49=AB^2+9-3AB,$ so $AB=8.$ By Ptolemy, $AC\cdot BD=AB\cdot CD+BC\cdot DA$ $=8\cdot3+3\cdot5=39,$ hence $BD=\tfrac{39}{7}.$ This is shorter than $AC=7.$

Thus, the correct answer is D.

20.

Points $P$ and $Q$ are chosen uniformly and independently at random on sides $\overline{AB}$ and $\overline{AC},$ respectively, of equilateral triangle $\triangle ABC.$ Which of the following intervals contains the probability that the area of $\triangle APQ$ is less than half the area of $\triangle ABC?$

$\left[\tfrac38,\tfrac12\right]$

$\left(\tfrac12,\tfrac23\right]$

$\left(\tfrac23,\tfrac34\right]$

$\left(\tfrac34,\tfrac78\right]$

$\left(\tfrac78,1\right]$

Answer: D

Concepts:geometric probability area ratio

Difficulty rating: 2100

Small Hint:

Let $x=\dfrac{AP}{AB}$ and $y=\dfrac{AQ}{AC}$ be uniform on $[0,1];$ the area ratio is $xy$

Big Hint:

Compute $P(xy\ge\tfrac12)=\displaystyle\int_{\frac{1}{2}}^{1}\left(1-\tfrac{1}{2x}\right)dx$ and subtract from $1$

Solution:

With $x=\tfrac{AP}{AB}$ and $y=\tfrac{AQ}{AC}$ uniform on $[0,1],$ the area ratio $\tfrac{[APQ]}{[ABC]}=xy.$ The complementary event $xy\ge\tfrac12$ requires $x\ge\tfrac12$ and $y\in[\tfrac{1}{2x},1],$ with probability $\begin{aligned} &\int_{\frac{1}{2}}^{1}\left(1-\frac{1}{2x}\right)dx \\ &=\frac12-\frac{\ln2}{2}\approx0.153. \end{aligned}$ Therefore $P(xy\lt\tfrac12)\approx1-0.153=0.847,$ which lies in $\left(\tfrac34,\tfrac78\right].$ Thus, the correct answer is D.

21.

Suppose that $a_1=2$ and the sequence $(a_n)$ satisfies the recurrence relation $\frac{a_n-1}{n-1}=\frac{a_{n-1}+1}{(n-1)+1}$ for all $n\ge2.$ What is the greatest integer less than or equal to $\sum_{n=1}^{100}a_n^2?$

$338{,}550$

$338{,}551$

$338{,}552$

$338{,}553$

$338{,}554$

Answer: B

Concepts:recursion sum of first n squares bounding to limit cases

Difficulty rating: 2130

Small Hint:

Compute a few terms and conjecture the closed form $a_n=n+\dfrac1n$

Big Hint:

Then $a_n^2=n^2+2+\dfrac1{n^2},$ and $\displaystyle\sum_{n=1}^{100}\dfrac1{n^2}$ lies strictly between $1$ and $2$

Solution:

The recurrence rearranges to $a_n=1+\tfrac{n-1}{n}(a_{n-1}+1).$ Computing early terms $2,\tfrac52,\tfrac{10}3,\tfrac{17}4,\ldots$ suggests $a_n=n+\tfrac1n.$ This follows by induction: substituting $a_{n-1}=n-1+\tfrac1{n-1}$ into the recurrence gives $a_n=1+\tfrac{n-1}{n}(n+\tfrac1{n-1})$ $=n+\tfrac1n.$ Then $a_n^2=n^2+2+\tfrac1{n^2},$ so $\begin{aligned} &\sum_{n=1}^{100}a_n^2 \\ &=\sum_{n=1}^{100}n^2+200 \\ &\quad {}+\sum_{n=1}^{100}\frac1{n^2} \\ &=338350+200+S, \end{aligned}$ where $S\gt1$ and $S\lt1+\int_1^\infty x^{-2}\,dx=2.$ Hence the sum is between $338551$ and $338552,$ and its floor is $338551.$ Thus, the correct answer is B.

22.

The figure below shows a dotted grid $8$ cells wide and $3$ cells tall consisting of $1''\times1''$ squares. Carl places $1$ -inch toothpicks along some of the sides of the squares to create a closed loop that does not intersect itself. The numbers in the cells indicate the number of sides of that square that are to be covered by toothpicks, and any number of toothpicks are allowed if no number is written. In how many ways can Carl place the toothpicks?

A dotted grid 8 cells wide and 3 cells tall, with a 1 written in each cell of the middle row.

$130$

$144$

$146$

$162$

$196$

Answer: C

Concepts:graph theory casework

Difficulty rating: 2370

Small Hint:

Each middle-row cell must touch exactly one toothpick; after the two turnaround columns are fixed, each interior cell’s toothpick is independently above or below it

Big Hint:

A loop crossing the middle uses all $8,$ the first $7,$ the last $7,$ or the middle $6$ columns; its interior columns independently bend above or below

Solution:

Each middle-row cell must touch exactly one toothpick. First consider loops that pass from one side of the middle strip to the other. The loop can span all $8$ columns, the first $7,$ the last $7,$ or the middle $6;$ a narrower span would leave an outer middle cell untouched.

Once the two ends are fixed, each interior middle cell independently has its one toothpick on its top or bottom side, and the rest of the non-self-intersecting loop is forced. The four cases therefore contribute $2^6,2^5,2^5,$ and $2^4$ loops. There are also exactly two loops that do not cross the middle strip: the horizontal rectangle running entirely along the top or entirely along the bottom. Hence the total is $2^6+2^5+2^5+2^4+2$ $=64+32+32+16+2$ $=146.$

Thus, the correct answer is C.

23.

What is the value of $\begin{aligned} &\tan^2\frac{\pi}{16}\cdot\tan^2\frac{3\pi}{16} \\ &\quad {}+\tan^2\frac{\pi}{16}\cdot\tan^2\frac{5\pi}{16} \\ &\quad {}+\tan^2\frac{3\pi}{16}\cdot\tan^2\frac{7\pi}{16} \\ &\quad {}+\tan^2\frac{5\pi}{16}\cdot\tan^2\frac{7\pi}{16}? \end{aligned}$

$28$

$68$

$70$

$72$

$84$

Answer: B

Concepts:trigonometric identity factoring

Difficulty rating: 2370

Small Hint:

Let $a,b,c,d=\tan^2\frac{\pi}{16},$ $\tan^2\frac{3\pi}{16},$ $\tan^2\frac{5\pi}{16},$ $\tan^2\frac{7\pi}{16};$ the sum factors as $(a+d)(b+c)$

Big Hint:

$\tan\frac{7\pi}{16}=\cot\frac{\pi}{16},$ and $\tan^2x+\cot^2x=\dfrac{4}{\sin^2 2x}-2$

Solution:

With $a=\tan^2\tfrac{\pi}{16},\$ $b=\tan^2\tfrac{3\pi}{16},\$ $c=\tan^2\tfrac{5\pi}{16},\$ $d=\tan^2\tfrac{7\pi}{16},$ the expression is $ab+ac+bd+cd$ $=(a+d)(b+c).$

Since $\tfrac{7\pi}{16}=\tfrac{\pi}{2}-\tfrac{\pi}{16},$ we have $d=\cot^2\tfrac{\pi}{16},$ so $a+d=\tan^2\tfrac{\pi}{16}+\cot^2\tfrac{\pi}{16}$ $=\tfrac{4}{\sin^2(\frac{\pi}{8})}-2$ $=14+8\sqrt2.$ Likewise $b+c=\tfrac{4}{\sin^2(\frac{3\pi}{8})}-2$ $=14-8\sqrt2.$ Their product is $14^2-(8\sqrt2)^2=196-128=68.$

Thus, the correct answer is B.

24.

A disphenoid is a tetrahedron whose triangular faces are congruent to one another. What is the least total surface area of a disphenoid whose faces are scalene triangles with integer side lengths?

$\sqrt3$

$3\sqrt{15}$

$15$

$15\sqrt7$

$24\sqrt6$

Answer: D

Concepts:3D geometry Heron’s Formula

Difficulty rating: 2520

Small Hint:

A disphenoid can be built from a triangle exactly when that triangle is acute; its surface is $4$ faces

Big Hint:

Seek the smallest-area acute scalene integer triangle; $(3,4,5)$ is right, so try $(4,5,6)$

Solution:

A disphenoid exists (as the tetrahedron formed by the face-plane midpoints of a box) exactly when the common face triangle is acute, and its total surface area is $4$ times one face’s area. Write the integer side lengths as $u\lt v\lt w.$ If $u\le3,$ then $w\ge v+1$ and $u^2+v^2\le(v+1)^2$ (with equality only for $(u,v,w)=(3,4,5)$ ), so the triangle is not acute. Thus $u\ge4.$

The triangle $(4,5,6)$ is acute because $4^2+5^2\gt6^2.$ It also has the least possible area. The largest angle of any acute triangle is at least $60^\circ,$ so a candidate with $(u,v)\ne(4,5)$ has area at least $\tfrac12uv\sin60^\circ$ $\ge\tfrac12(4)(6)\tfrac{\sqrt3}{2}=6\sqrt3,$ which is greater than $\tfrac{15\sqrt7}{4},$ the area of $(4,5,6).$

By Heron’s formula with $s=\tfrac{15}2,$ that area is $\sqrt{\tfrac{15}2\cdot\tfrac72\cdot\tfrac52\cdot\tfrac32}=\tfrac{15\sqrt7}{4}.$ The total surface area is $4\cdot\tfrac{15\sqrt7}{4}=15\sqrt7.$

Thus, the correct answer is D.

25.

A graph is symmetric about a line if the graph remains unchanged after reflection in that line. For how many quadruples of integers $(a,b,c,d),$ where $|a|,$ $|b|,$ $|c|,$ $|d|\le5$ and $c$ and $d$ are not both $0,$ is the graph of $y=\frac{ax+b}{cx+d}$ symmetric about the line $y=x?$

$1282$

$1292$

$1310$

$1320$

$1330$

Answer: B

Concepts:function casework

Difficulty rating: 2720

Small Hint:

Reflecting $y=f(x)$ over $y=x$ gives its inverse, so symmetry means $f$ is its own inverse

Big Hint:

A map $\dfrac{ax+b}{cx+d}$ is an involution exactly when $a+d=0$ (and it is nondegenerate); also count the identity $y=x$

Solution:

Reflecting the graph of $y=f(x)$ over $y=x$ produces the graph of its inverse, so the graph is symmetric about $y=x$ exactly when $f$ equals its own inverse. For $f(x)=\tfrac{ax+b}{cx+d}$ this happens in two ways: when $a+d=0$ with $ad-bc\ne0$ (a genuine involution, including the slope $-1$ lines when $c=0$ ), or when $f$ is the identity $y=x$ ( $b=c=0,\ a=d\ne0$ ).

For $a+d=0,$ set $d=-a;$ the determinant $-a^2-bc$ must be nonzero, so we need $a^2+bc\ne0,$ together with $(c,d)\ne(0,0).$ When $a=0,$ both $b$ and $c$ must be nonzero, giving $10^2=100$ choices. For each nonzero $a,$ start with $11^2=121$ choices of $(b,c).$ If $|a|=1,3,4,$ or $5,$ exactly $2$ pairs satisfy $bc=-a^2;$ if $|a|=2,$ exactly $6$ pairs do. Thus the genuine involutions number $\begin{gathered} 100+8(121-2)+2(121-6)\\ {}=1282. \end{gathered}$ The identity case adds $10$ more ( $a=d\in\{\pm1,\ldots,\pm5\}$ ), for a total of $1282+10=1292.$

Thus, the correct answer is B.