2010 AMC 12A Solutions | LIVE by Po-Shen Loh

1.

What is $(20-(2010-201))$ $+(2010-(201-20))?$

$-4020$

$0$

$40$

$401$

$4020$

Concepts:whole number operations order of operations

Difficulty rating: 770

Small Hint:

Distribute the negative signs and look for terms that cancel

Big Hint:

After the $2010$ and $201$ terms cancel, only the two $20$ s remain

Solution:

Distributing the negative signs gives $\begin{aligned} &(20-2010+201) \\ &\quad {}+(2010-201+20)=40. \end{aligned}$

Thus, C is the correct answer.

2.

A ferry boat shuttles tourists to an island every hour starting at $10$ am until its last trip, which starts at $3$ pm. One day the boat captain notes that on the $10$ am trip there were $100$ tourists on the ferry boat, and that on each successive trip, the number of tourists was $1$ fewer than on the previous trip. How many tourists did the ferry take to the island that day?

$585$

$594$

$672$

$679$

$694$

Concepts:arithmetic sequence summation

Difficulty rating: 970

Small Hint:

Count how many trips run from $10$ am through $3$ pm

Big Hint:

The trip counts are $100,99,\ldots,95;$ add them

Solution:

The ferry makes $6$ trips: at $10,$ $11,$ $12,$ $1,$ $2,$ and $3.$

The numbers of tourists are $100,99,98,97,96,95,$ so the total is $\begin{gathered} 6\cdot100-(1+2+3+4+5) \\ =600-15 \\ =585. \end{gathered}$

Thus, A is the correct answer.

3.

Rectangle $ABCD,$ pictured below, shares $50\%$ of its area with square $EFGH.$ Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD.$ What is $\dfrac{AB}{AD}?$

$4$

$5$

$6$

$8$

$10$

Concepts:area ratio percentage

Difficulty rating: 1120

Small Hint:

Let $s$ be the side length of the square; the shaded overlap has width $s$ and height $AD$

Big Hint:

Half the rectangle’s area equals the overlap, giving $AB=2s;$ one fifth of the square’s area gives $AD=\tfrac{s}{5}$

Solution:

Let $s$ be the side length of square $EFGH.$ The shaded overlap has width $s$ and height $AD,$ so its area is $s\cdot AD.$

Because the overlap is $50\%$ of the rectangle, $s\cdot AD=\tfrac12\,AB\cdot AD,$ so $AB=2s.$ Because it is $20\%$ of the square, $s\cdot AD=\tfrac15 s^2,$ so $AD=\tfrac{s}{5}.$

Therefore $\dfrac{AB}{AD}=\dfrac{2s}{\frac{s}{5}}=10.$

Thus, E is the correct answer.

4.

If $x\lt0,$ then which of the following must be positive?

$\dfrac{x}{|x|}$

$-x^2$

$-2^x$

$-x^{-1}$

$\sqrt[3]{x}$

Concepts:inequality exponent absolute value

Difficulty rating: 1070

Small Hint:

Test each choice with a negative value such as $x=-1$

Big Hint:

Rewrite $-x^{-1}$ as $-\dfrac1x;$ dividing into a negative number flips the sign

Solution:

Choice (D) is $-x^{-1}=-\dfrac1x.$ When $x\lt0,$ $\dfrac1x\lt0,$ so $-\dfrac1x\gt0.$

Testing $x=-1$ shows the other choices need not be positive: $\dfrac{x}{|x|}=-1,$ $-x^2=-1,$ $-2^x=-\tfrac12,$ and $\sqrt[3]{x}=-1.$

Thus, D is the correct answer.

5.

Halfway through a $100$ -shot archery tournament, Chelsea leads by $50$ points. For each shot a bullseye scores $10$ points, with other possible scores being $8,$ $4,$ $2,$ and $0$ points. Chelsea always scores at least $4$ points on each shot. If Chelsea’s next $n$ shots are bullseyes she will be guaranteed victory. What is the minimum value for $n?$

$38$

$40$

$42$

$44$

$46$

Concepts:inequality bounding to limit cases

Difficulty rating: 1350

Small Hint:

Chelsea’s opponent can score at most $10$ points on each of the remaining $50$ shots

Big Hint:

With $n$ bullseyes and at least $4$ points on the other $50-n$ shots, require $10n+4(50-n)\gt 500-50$

Solution:

The opponent can score at most $50\cdot10=500$ on the last $50$ shots. Since Chelsea leads by $50,$ she must score more than $500-50=450$ points on her remaining shots to guarantee victory.

Her $n$ bullseyes give $10n$ points, and her other $50-n$ shots give at least $4(50-n)$ points, so $10n+4(50-n)\gt450.$ This simplifies to $6n\gt250,$ i.e. $n\gt41\tfrac23.$

Therefore Chelsea needs at least $42$ bullseyes.

Thus, C is the correct answer.

6.

A palindrome, such as $83438,$ is a number that remains the same when its digits are reversed. The numbers $x$ and $x + 32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x?$

$20$

$21$

$22$

$23$

$24$

Concepts:palindrome digits bounding to limit cases

Difficulty rating: 1280

Small Hint:

Bound the four-digit palindrome between $1000$ and $1031$

Big Hint:

Find the only palindrome in that interval, then subtract $32$

Solution:

Note that $x$ is at most $999.$ This means that $x + 32$ has a maximum of $1031.$

Similarly, we have that the minimum value of $x + 32$ is $1000.$

The only palindrome in this range is $1001,$ so this is what $x + 32$ equals.

Thus $x=1001-32=969,$ whose digit sum is $9+6+9=24.$

Thus, E is the correct answer.

7.

Logan is constructing a scaled model of his town. The city’s water tower stands $40$ meters high, and the top portion is a sphere that holds $100{,}000$ liters of water. Logan’s miniature water tower holds $0.1$ liters. How tall, in meters, should Logan make his tower?

$0.04$

$\dfrac{0.4}{\pi}$

$0.4$

$\dfrac{4}{\pi}$

$4$

Concepts:power scaling of length, area, and volume volume

Difficulty rating: 1420

Small Hint:

Volume scale is the cube of length scale

Big Hint:

Compare $0.1$ liters to $100000$ liters, then take a cube root

Solution:

The model-to-original volume ratio is $\dfrac{0.1}{100{,}000}=\dfrac{1}{1{,}000{,}000}.$ Lengths scale by the cube root, so the model-to-original height ratio is $\dfrac1{100}.$ Therefore the model should be $\dfrac{40}{100}=0.4$ meters tall.

Thus, C is the correct answer.

8.

Triangle $ABC$ has $AB=2 \cdot AC.$ Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC},$ respectively, such that $\angle BAE = \angle ACD.$ Let $F$ be the intersection of segments $AE$ and $CD,$ and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB?$

$60^\circ$

$75^\circ$

$90^\circ$

$105^\circ$

$120^\circ$

Concepts:angle chasing equilateral triangle special right triangle

Difficulty rating: 1660

Small Hint:

Let $\angle BAE=\angle ACD=x$

Big Hint:

Use $\triangle CFE$ to find $\angle AFC,$ then angle-chase $\angle BAC$

Solution:

Let $\angle BAE=\angle ACD=x.$ Because $\triangle CFE$ is equilateral, $\angle CFE=60^\circ.$ Rays $FA$ and $FE$ are opposite, so $\angle AFC=120^\circ.$

In $\triangle AFC,$ we get $\angle FAC=180^\circ-120^\circ-x$ $=60^\circ-x.$ Since $F$ lies on $AE,$ this is $\angle EAC.$ Hence $\angle BAC=x+(60^\circ-x)=60^\circ.$

Put $AC=s,$ so $AB=2s.$ The Law of Cosines gives $\begin{aligned} BC^2 &= s^2+(2s)^2 \\ &\quad-2(s)(2s)\cos60^\circ \\ &=3s^2. \end{aligned}$ Thus the side lengths are in the ratio $1:\sqrt3:2,$ with $AB$ the hypotenuse, so $\angle ACB=90^\circ.$

Thus, C is the correct answer.

9.

A solid cube has side length $3$ inches. A $2$ -inch by $2$ -inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?

$7$

$8$

$10$

$12$

$15$

Concepts:volume inclusion-exclusion cube geometry

Difficulty rating: 1790

Small Hint:

Use inclusion-exclusion for the three rectangular holes

Big Hint:

The three $2\times2\times3$ holes overlap in the central $2\times2\times2$ cube

Solution:

Note that all the cut out solids intersect in the middle of the cube.

This region of intersection is a cube with side length $2.$ Then the volume of the cutout region is $\begin{align*}3 \cdot 2 \cdot 2 \cdot 3 - 2 \cdot 2^3 &= 36 - 16 \\&= 20.\end{align*}$

We have to subtract out the center region twice since it is included in all $3$ regions.

The remaining volume is then $3^3 - 20 = 27 - 20 = 7.$

Thus, A is the correct answer.

10.

The first four terms of an arithmetic sequence are $p,$ $9,$ $3p-q,$ and $3p+q.$ What is the $2010$ th term of this sequence?

$8041$

$8043$

$8045$

$8047$

$8049$

Concepts:arithmetic sequence system of equations

Difficulty rating: 1410

Small Hint:

The common difference equals $(3p+q)-(3p-q)=2q$

Big Hint:

Also $9-p=2q$ and $(3p-q)-9=2q;$ solve for $p$ and $q$

Solution:

Consecutive terms differ by a common difference $d.$ From the last two terms, $d=(3p+q)-(3p-q)=2q.$

From the first two terms, $9-p=d=2q,$ and from the second and third, $(3p-q)-9=d=2q.$ Solving this system gives $p=5,$ $q=2,$ and $d=4.$

The $2010$ th term is $\begin{aligned} p+2009d &= 5+2009\cdot4 \\ &= 8041. \end{aligned}$

Thus, A is the correct answer.

11.

The solution of the equation $7^{x+7}=8^x$ can be expressed in the form $x=\log_b 7^7.$ What is $b?$

$\dfrac{7}{15}$

$\dfrac{7}{8}$

$\dfrac{8}{7}$

$\dfrac{15}{8}$

$\dfrac{15}{7}$

Concepts:logarithm exponent

Difficulty rating: 1510

Small Hint:

$x=\log_b 7^7$ means $b^x=7^7$

Big Hint:

Multiply to combine the bases: $(7b)^x=7^x\cdot b^x=7^{x+7}$

Solution:

Since $x=\log_b 7^7,$ we have $b^x=7^7.$

Then $\begin{aligned} (7b)^x=7^x\cdot b^x &= 7^x\cdot7^7 \\ &= 7^{x+7}=8^x. \end{aligned}$

Because $x\gt0,$ it follows that $7b=8,$ so $b=\dfrac{8}{7}.$

Thus, C is the correct answer.

12.

In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.

Brian: “Mike and I are different species.”

Chris: “LeRoy is a frog.”

LeRoy: “Chris is a frog.”

Mike: “Of the four of us, at least two are toads.”

How many of these four amphibians are frogs?

$0$

$1$

$2$

$3$

$4$

Concepts:truth-tellers and liars logical deduction

Difficulty rating: 1540

Small Hint:

Chris and LeRoy cannot have the same species

Big Hint:

If Brian were a toad, Mike’s statement would create a contradiction

Solution:

Chris and LeRoy are of opposite species: if Chris is a toad, his statement makes LeRoy a frog; if Chris is a frog, his false statement makes LeRoy a toad. Thus exactly one of them is a toad.

If Brian is a toad, his true statement makes Mike a frog. If Brian is a frog, his false statement says that he and Mike are the same species, again making Mike a frog. Therefore Mike is always a frog.

Mike’s statement is therefore false, so there are fewer than two toads. Chris and LeRoy already supply exactly one toad, forcing Brian to be a frog. Hence there is one toad and $3$ frogs.

Thus, D is the correct answer.

13.

For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy=k$ not intersect?

$0$

$1$

$2$

$4$

$8$

Concepts:circle hyperbola casework

Difficulty rating: 1590

Small Hint:

For $k\ne0,$ $x^2+y^2=k^2$ is a circle of radius $|k|$ and $xy=k$ is a hyperbola

Big Hint:

The hyperbola’s points closest to the origin are at distance $\sqrt{2|k|};$ compare with the radius $|k|$

Solution:

For $k=0,$ the graph of $x^2+y^2=0$ is the single point $(0,0)$ and $xy=0$ is the two axes, which meet at the origin, so the graphs intersect.

For $k\ne0,$ the circle has radius $|k|,$ and the hyperbola $xy=k$ has its two vertices nearest the origin at distance $\sqrt{2|k|}.$ The graphs meet exactly when $|k|\ge\sqrt{2|k|},$ that is $|k|\ge2.$

So they fail to intersect only when $|k|=1,$ namely $k=1$ and $k=-1,$ giving $2$ values.

Thus, C is the correct answer.

14.

Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3,$ and $DC = 8.$ What is the smallest possible value of the perimeter?

$30$

$33$

$35$

$36$

$37$

Concepts:angle bisector theorem triangle inequality optimization

Difficulty rating: 1600

Small Hint:

Use the Angle Bisector Theorem

Big Hint:

$AB:BC=3:8,$ and $AC=11$

Solution:

Using the Angle Bisector Theorem, we have that $\dfrac{AB}{3} = \dfrac{BC}{8}$ $AB = \dfrac{3}{8} BC.$

For $AB$ and $BC$ to be integers, we must have that $BC$ is a multiple of $8.$

To minimize the perimeter, we can set $BC = 8$ and $AB = 3.$ This, however, makes the triangle degenerate.

$BC$ must then be $16$ and $AB = 6.$ Since $AC = AD + DC = 11,$ the perimeter is $16 + 6 + 11 = 33.$

Thus, B is the correct answer.

15.

A coin is altered so that the probability that it lands on heads is less than $\dfrac12,$ and when the coin is flipped four times, the probability of an equal number of heads and tails is $\dfrac{1}{6}.$ What is the probability that the coin lands on heads?

$\dfrac{\sqrt{15}-3}{6}$

$\dfrac{6-\sqrt{6\sqrt6+2}}{12}$

$\dfrac{\sqrt2-1}{2}$

$\dfrac{3-\sqrt3}{6}$

$\dfrac{\sqrt3-1}{2}$

Concepts:binomial probability quadratic

Difficulty rating: 1650

Small Hint:

The chance of two heads and two tails is $\binom{4}{2}p^2(1-p)^2$

Big Hint:

This reduces to $p(1-p)=\dfrac16;$ solve the quadratic and take the root with $p\lt\dfrac12$

Solution:

Let $p$ be the probability of heads. The chance of two heads and two tails in four flips is $\begin{aligned} \binom{4}{2}p^2(1-p)^2 &= 6p^2(1-p)^2 \\ &= \frac16. \end{aligned}$

Thus $p^2(1-p)^2=\dfrac{1}{36},$ so $p(1-p)=\dfrac16.$

This gives $6p^2-6p+1=0,$ so $p=\dfrac{3\pm\sqrt3}{6}.$ Since $p\lt\dfrac12,$ we take $p=\dfrac{3-\sqrt3}{6}.$

Thus, D is the correct answer.

16.

Bernardo randomly picks $3$ distinct numbers from the set $\{1,2,3,4,5,6,7,8,9\}$ and arranges them in descending order to form a $3$ -digit number. Silvia randomly picks $3$ distinct numbers from the set $\{1,2,3,4,5,6,7,8\}$ and also arranges them in descending order to form a $3$ -digit number. What is the probability that Bernardo’s number is larger than Silvia’s number?

$\dfrac{47}{72}$

$\dfrac{37}{56}$

$\dfrac{2}{3}$

$\dfrac{49}{72}$

$\dfrac{39}{56}$

Concepts:basic probability casework symmetry

Difficulty rating: 1900

Small Hint:

Separate cases according to whether Bernardo picks $9$

Big Hint:

Without a $9,$ the two numbers are symmetric except when the chosen sets match

Solution:

There are two cases: Bernardo picks a $9$ or he doesn’t.

Case $1:$ Bernardo picks a $9$

Since a number is fixed, there are $\binom{8}{2} = 28$ ways to choose the other two numbers.

There are a total of $\binom{9}{3} = 84$ ways to pick all three numbers. The probability is then $\dfrac{28}{84} = \dfrac{1}{3}.$

Note that if Bernardo picks a $9,$ he automatically has a greater number than Silvia.

This means that Bernardo always wins in this case.

Case $2:$ Bernardo doesn’t pick a $9$

There is a $1 - \frac{1}{3} = \frac{2}{3}$ chance of this happening. Since both people are choosing from the same numbers, they have an equal chance of winning.

We still need to find the probability that the numbers are the same. There is a $\dfrac{1}{\binom{8}{3}} = \dfrac{1}{56}$ chance that Silvia chooses the same numbers as Bernardo. The probability that Bernardo gets a higher number is then $\dfrac{1 - \frac{1}{56}}{2} = \dfrac{55}{112}.$

The total probability of Bernardo getting a higher number is then $\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{55}{112} = \dfrac{37}{56}.$

Thus, B is the correct answer.

17.

Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r.$ The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r?$

$\dfrac{4\sqrt{3}}{3}$

$\dfrac{10}{3}$

$4$

$\dfrac{17}{4}$

$6$

Concepts:equiangular polygon law of cosines Vieta’s Formulas

Difficulty rating: 1960

Small Hint:

Split the hexagon into $\triangle ACE$ and three corner triangles

Big Hint:

Express both areas using $r^2+r+1$ and $r$

Solution:

Note that $\triangle ACE$ is equilateral. Using the Law of Cosines in $\triangle ABC,$ we get $AC^2=r^2+1^2-2r\cos120^\circ$ $=r^2+r+1.$

The area of $\triangle ACE$ is then $\dfrac{\sqrt3}{4} (r^2 + r + 1).$

The three corner triangles $\triangle ABC,$ $\triangle CDE,$ and $\triangle EFA$ each have area $\frac12\cdot1\cdot r\cdot\sin120^\circ=\frac{r\sqrt3}{4}.$

Thus the hexagon has area $\dfrac{\sqrt3}{4}(r^2+r+1)$ $+3\cdot\dfrac{r\sqrt3}{4}$ $=\dfrac{\sqrt3}{4}(r^2+4r+1).$

The condition $[ACE]=70\%\cdot[ABCDEF]$ gives $r^2+r+1=\dfrac{7}{10}(r^2+4r+1),$ so $r^2-6r+1=0.$

By Vieta’s formulas, the sum of the possible values of $r$ is $6.$

Thus, E is the correct answer.

18.

A $16$ -step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$ -coordinate or the $y$ -coordinate by $1.$ How many such paths stay outside or on the boundary of the square $-2\le x\le2,$ $-2\le y\le2$ at each step?

$92$

$144$

$1568$

$1698$

$12{,}800$

Concepts:lattice paths combinations symmetry

Difficulty rating: 1880

Small Hint:

Each step increases $x+y$ by $1,$ so the path meets the line $x+y=0$ at exactly one lattice point

Big Hint:

To avoid the square’s interior, that point is $(\pm2,\mp2),(\pm3,\mp3),$ or $(\pm4,\mp4);$ count paths through each

Solution:

Every step increases $x+y$ by $1,$ which runs from $-8$ to $8,$ so each path passes through exactly one lattice point with $x+y=0.$

To stay out of the open square, that point $(t,-t)$ must have $|t|\ge2,$ so it is one of $(\pm2,\mp2),(\pm3,\mp3),(\pm4,\mp4).$

By symmetry consider the three points $(-4,4),(-3,3),(-2,2)$ and double. The number of paths from $(-4,-4)$ to $(-(4-j),4-j)$ is $\binom{8}{j},$ and the number continuing on to $(4,4)$ is also $\binom{8}{j}.$

Therefore the total is $\begin{gathered} 2\left(\binom80^2+\binom81^2+\binom82^2\right) \\ =2(1+64+784) \\ =1698. \end{gathered}$

Thus, D is the correct answer.

19.

Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010,$ the box in the $k$ th position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n) \lt \dfrac{1}{2010}?$

$45$

$63$

$64$

$201$

$1005$

Concepts:telescoping basic probability inequality

Difficulty rating: 2240

Small Hint:

The first $n-1$ draws must be white, then the $n$ th draw red

Big Hint:

The white probabilities telescope: $\dfrac12\cdot\dfrac23\cdots\dfrac{n-1}{n}$

Solution:

Since there are $k + 1$ marbles in the $k$ th box, there is a $\dfrac{k}{k + 1}$ chance Isabella draws a white marble from it.

The probability of drawing a red marble is then $\dfrac{1}{k + 1}.$ To stop after drawing the $n$ th marble, the first $n - 1$ marbles must have been white.

This happens with a probability of $\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}.$

Note that all the numerators cancel with the adjacent denominator, which means that this expression reduces to $\dfrac{1}{n(n + 1)}.$

We have to find the smallest $n$ such that $\dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010}$ $n(n + 1) \gt 2010.$

Since $44\cdot45=1980\lt2010$ but $45\cdot46=2070\gt2010,$ the smallest such $n$ is $45.$

Thus, A is the correct answer.

20.

Arithmetic sequences $(a_n)$ and $(b_n)$ have integer terms with $a_1=b_1=1\lt a_2\le b_2$ and $a_nb_n=2010$ for some $n.$ What is the largest possible value of $n?$

$2$

$3$

$8$

$288$

$2009$

Concepts:arithmetic sequence divisibility greatest common divisor

Difficulty rating: 1950

Small Hint:

Write $a_n=1+(n-1)d_1$ and $b_n=1+(n-1)d_2,$ so $n-1$ divides both $a_n-1$ and $b_n-1$

Big Hint:

Check each factor pair of $2010$ with $a_n\le b_n;$ then $n-1$ must divide $\gcd(a_n-1,b_n-1)$

Solution:

Since $a_n=1+(n-1)d_1$ and $b_n=1+(n-1)d_2$ for integers $d_1,d_2,$ the value $n-1$ divides both $a_n-1$ and $b_n-1,$ hence divides $\gcd(a_n-1,b_n-1).$

The factor pairs of $2010$ with $2\le a_n\le b_n$ are $(2,1005),$ $(3,670),$ $(5,402),$ $(6,335),$ $(10,201),$ $(15,134),$ and $(30,67).$

For every pair except $(15,134),$ the numbers $a_n-1$ and $b_n-1$ are relatively prime, forcing $n=2.$ For $(15,134),$ $\gcd(14,133)=7,$ so $n-1$ can equal $7,$ giving $n=8.$

The sequences $a_n=2n-1$ and $b_n=19n-18$ realize this, so the largest value is $8.$

Thus, C is the correct answer.

21.

The graph of $y=x^6-10x^5$ $+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x,$ where the graph and the line intersect. What is the largest of those values?

$4$

$5$

$6$

$7$

$8$

Concepts:polynomial factoring completing the square

Difficulty rating: 2000

Small Hint:

The graph minus the line is nonnegative with three double roots, so it equals $\big((x-p)(x-q)(x-r)\big)^2$

Big Hint:

Match the coefficients of $x^5,x^4,x^3$ to determine the cubic, then factor it

Solution:

Let $f(x)$ be the graph minus the line. It is nonnegative and vanishes at three points, each a double root, so $f(x)=\big(x^3-Ax^2+Bx-C\big)^2.$

Matching coefficients gives $-2A=-10\Rightarrow A=5,$ then $A^2+2B=29\Rightarrow B=2,$ then $-2C-2AB=-4\Rightarrow C=-8.$

Thus the cubic is $x^3-5x^2+2x+8$ $=(x+1)(x-2)(x-4),$ with roots $-1,2,$ and $4.$ The largest is $4.$

Thus, A is the correct answer.

22.

What is the minimum value of $\begin{gathered} f(x) = |x-1|+|2x-1| \\ {}+|3x-1|+\cdots+|119x-1|? \end{gathered}$

$49$

$50$

$51$

$52$

$53$

Concepts:absolute value optimization summation

Difficulty rating: 2000

Small Hint:

$f$ is piecewise linear with corners at $x=\tfrac1k;$ the minimum is where its slope turns from negative to positive

Big Hint:

On $\left[\tfrac1m,\tfrac1{m-1}\right]$ the slope is $7140-(m-1)m;$ find where it vanishes

Solution:

The function $f$ is piecewise linear with breakpoints at $x=\tfrac1k.$ On the interval $\left[\tfrac1m,\tfrac1{m-1}\right]$ its slope is $\begin{aligned} &\sum_{k=m}^{119}k \\ &\quad {}-\sum_{k=1}^{m-1}k=7140-(m-1)m, \end{aligned}$ where $7140=\tfrac{119\cdot120}{2}.$

This slope is zero when $(m-1)m=7140,$ i.e. $m=85,$ so the minimum occurs at the right endpoint $x=\tfrac1{84}.$

There, terms with $k\le84$ contribute $\tfrac{84-k}{84}$ and terms with $k\ge85$ contribute $\tfrac{k-84}{84},$ so $\begin{aligned} f\left(\tfrac1{84}\right) &= \frac{3486}{84}+\frac{630}{84} \\ &= 41.5+7.5=49. \end{aligned}$

Thus, A is the correct answer.

23.

The number obtained from the last two nonzero digits of $90!$ is equal to $n.$ What is $n?$

$12$

$32$

$48$

$52$

$68$

Concepts:modular arithmetic Chinese Remainder Theorem trailing zeros

Difficulty rating: 2390

Small Hint:

Remove the factors making trailing zeroes, then work modulo $100$

Big Hint:

Use congruences modulo $4$ and modulo $25$ to pin down the last two digits

Solution:

The number of trailing zeroes in $90!$ is $\left\lfloor\dfrac{90}{5}\right\rfloor+\left\lfloor\dfrac{90}{25}\right\rfloor=21.$ Let $N=\dfrac{90!}{10^{21}}.$

There are still more than two factors of $2$ left after removing $10^{21},$ so $N\equiv0 \pmod4.$

Let $A$ be the product of factors of $90!$ not divisible by $5,$ and let $B$ be the product of the factors divisible by $5.$ Each block $(5j+1)(5j+2)(5j+3)(5j+4)$ is $24\equiv-1\pmod{25},$ and there are $18$ blocks, so $A\equiv1\pmod{25}.$

After removing the $21$ factors of $5$ from $B,$ the remaining factors can be grouped as $\begin{aligned} \dfrac{B}{5^{21}}={}&(1\cdot2\cdot3\cdot4) \\ &\cdot(6\cdot7\cdot8\cdot9) \\ &\cdot(11\cdot12\cdot13\cdot14) \\ &\cdot(16\cdot17\cdot18)(1\cdot2\cdot3) \\ &\equiv-1\pmod{25}. \end{aligned}$

Therefore $\dfrac{90!}{5^{21}}\equiv-1\pmod{25}.$ Since $2^{21}\equiv2\pmod{25}$ and the inverse of $2$ modulo $25$ is $13,$ we get $N\equiv-13\equiv12\pmod{25}.$

The number congruent to $0\pmod4$ and $12\pmod{25}$ is $12\pmod{100},$ so the last two nonzero digits form $12.$

Thus, A is the correct answer.

24.

Let $\begin{gathered} f(x)=\log_{10}\big(\sin(\pi x) \\ {}\cdot\sin(2\pi x)\cdot\sin(3\pi x) \\ {}\cdot\sin(4\pi x)\cdot\sin(5\pi x) \\ {}\cdot\sin(6\pi x)\cdot\sin(7\pi x) \\ {}\cdot\sin(8\pi x)\big). \end{gathered}$ The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n?$

$2$

$12$

$18$

$22$

$36$

Concepts:trigonometry symmetry casework

Difficulty rating: 2460

Small Hint:

The domain is where $g(x)=\prod_{k=1}^8\sin(k\pi x)\gt0;$ $g$ is symmetric about $x=\tfrac12$

Big Hint:

Count the zeros $\tfrac{k}{n}$ in $\big(0,\tfrac12\big)$ and note the sign fails to flip at $\tfrac14$ and $\tfrac13$

Solution:

Let $g(x)=\prod_{k=1}^8\sin(k\pi x);$ the domain of $f$ is where $g(x)\gt0.$ Since $\sin(k\pi(1-x))$ $=(-1)^{k+1}\sin(k\pi x)$ and $\sum_{k=1}^8(k+1)$ is even, $g(1-x)=g(x),$ so it suffices to study $\big(0,\tfrac12\big)$ and double.

In $\big(0,\tfrac12\big)$ the zeros of $g$ are the fractions $\tfrac{k}{n}$ with $2\le n\le8,$ $1\le k\lt\tfrac n2,$ and $\gcd(k,n)=1.$ For $n=2,\ldots,8$ there are $0,1,1,2,1,3,2$ of them, totaling $10.$

These $10$ zeros split $\big(0,\tfrac12\big)$ into $11$ subintervals on which $g$ has constant sign. Near $0$ every factor is positive, so $g\gt0$ there, and the sign flips at each zero except $\tfrac14=\tfrac28$ and $\tfrac13=\tfrac26,$ where an even number of factors vanish.

Tracking the signs, exactly $6$ of the $11$ subintervals have $g\gt0.$ By symmetry there are $6$ more in $\big(\tfrac12,1\big),$ so $n=12.$

Thus, B is the correct answer.

25.

Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to $32?$

$560$

$564$

$568$

$1498$

$2255$

Concepts:Burnside’s Lemma stars and bars cyclic quadrilateral

Difficulty rating: 2520

Small Hint:

A convex cyclic quadrilateral with given side lengths in a given cyclic order is unique; each side is at most $15$

Big Hint:

Count ordered quadruples summing to $32,$ then quotient by cyclic rotation using Burnside’s lemma

Solution:

A convex cyclic quadrilateral is determined up to rotation and translation by its cyclic sequence of side lengths, and it exists exactly when the largest side is less than the sum of the others. With perimeter $32,$ this means each side is at most $15.$

First count ordered quadruples $(a,b,c,d)$ of positive integers with $a+b+c+d=32$ and each entry at most $15.$ Without the upper bound there are $\binom{31}{3}=4495;$ removing those with some entry at least $16$ subtracts $4\binom{16}{3}=2240,$ leaving $2255.$

Rotations of the quadrilateral correspond to cyclic permutations of $(a,b,c,d).$ By Burnside’s lemma the number of distinct quadrilaterals is $\frac{1}{4}\big(2255+f_1+f_2+f_3\big),$ where $f_i$ counts quadruples fixed by rotating $i$ positions.

A one- or three-step rotation fixes only $(8,8,8,8),$ so $f_1=f_3=1.$ A two-step rotation fixes $(a,b,a,b)$ with $a+b=16$ and $1\le a,b\le15,$ giving $f_2=15.$

Hence the count is $\begin{aligned} \frac{1}{4}(2255+1+15+1) &= \frac{2272}{4} \\ &= 568. \end{aligned}$

Thus, C is the correct answer.