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

1.

Makayla attended two meetings during her $9$ -hour work day. The first meeting took $45$ minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?

$15$

$20$

$25$

$30$

$35$

Concepts:percentage unit conversion

Difficulty rating: 800

Small Hint:

Convert the work day to minutes: $9$ hours is $540$ minutes

Big Hint:

The two meetings together last $45+2\cdot45$ minutes

Solution:

The two meetings lasted $45+90=135$ minutes, and the work day is $9\cdot60=540$ minutes.

The fraction of the day spent in meetings is $\frac{135}{540}=\frac14=25\%.$

Thus, the correct answer is C.

2.

A big L is formed as shown. What is its area?

$22$

$24$

$26$

$28$

$30$

Concepts:area decomposition rectangle

Difficulty rating: 880

Small Hint:

Split the L into two rectangles

Big Hint:

The vertical bar is $2\times8,$ and the foot adds a $2\times3$ piece since $5-2=3$

Solution:

The region splits into an $8\times2$ vertical rectangle and a $2\times3$ horizontal foot, whose width is $5-2=3.$

The total area is $8\cdot2+3\cdot2=16+6=22.$

Thus, the correct answer is A.

3.

A ticket to a school play costs $x$ dollars, where $x$ is a whole number. A group of $9$ th graders buys tickets costing a total of $\$48,$ and a group of $10$ th graders buys tickets costing a total of $\$64.$ How many values for $x$ are possible?

$1$

$2$

$3$

$4$

$5$

Concepts:greatest common divisor factor counting

Difficulty rating: 1010

Small Hint:

$x$ must divide both $48$ and $64$

Big Hint:

Count the common divisors of $48$ and $64$

Solution:

The price $x$ must divide both totals, so $x$ is a common divisor of $48$ and $64.$

Since $\gcd(48,64)=16,$ the common divisors are $1, 2, 4, 8,$ and $16.$ There are $5$ possible values.

Thus, the correct answer is E.

4.

A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?

$2$

$3$

$4$

$5$

$6$

Concepts:date and time casework

Difficulty rating: 1240

Small Hint:

$31=4\cdot7+3,$ so exactly three weekdays occur five times

Big Hint:

Those three days are the first three days of the month; check when Monday and Wednesday match

Solution:

Since $31=4\cdot7+3,$ the first three days of the month each occur five times, and the other four days occur four times.

Mondays and Wednesdays are equal in number exactly when both fall in the five-time group or both fall in the four-time group.

If the first day is Monday, the five-time days are Mon, Tue, Wed (both appear five times). If the first day is Thursday or Friday, the five-time days miss both Monday and Wednesday (both appear four times). Every other starting day includes exactly one of Monday or Wednesday.

So the first day can be Monday, Thursday, or Friday, giving $3$ possibilities.

Thus, the correct answer is B.

5.

Lucky Larry’s teacher asked him to substitute numbers for $a,$ $b,$ $c,$ $d,$ and $e$ in the expression $a-(b-(c-(d+e)))$ and evaluate the result. Larry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for $a,$ $b,$ $c,$ and $d$ were $1,$ $2,$ $3,$ and $4,$ respectively. What number did Larry substitute for $e?$

$-5$

$-3$

$0$

$3$

$5$

Concepts:order of operations linear equation

Difficulty rating: 1100

Small Hint:

Expand the correct expression to $a-b+c-d-e$

Big Hint:

Larry instead computed $a-b-c-d+e;$ set the two equal

Solution:

The correct value is $a-(b-(c-(d+e)))=$ $a-b+c-d-e.$ With $a, b, c, d=1, 2, 3, 4,$ this equals $1-2+3-4-e=-2-e.$

Larry dropped the parentheses and computed $1-2-3-4+e=-8+e.$

Setting $-2-e=-8+e$ gives $2e=6,$ so $e=3.$

Thus, the correct answer is D.

6.

At the beginning of the school year, $50\%$ of all students in Mr. Wells’ math class answered “Yes” to the question “Do you love math”, and $50\%$ answered “No.” At the end of the school year, $70\%$ answered “Yes” and $30\%$ answered “No.” Altogether, $x\%$ of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of $x?$

$0$

$20$

$40$

$60$

$80$

Concepts:percentage extremal argument

Difficulty rating: 1410

Small Hint:

Assume $100$ students; the “Yes” count rises from $50$ to $70$

Big Hint:

At least $20$ must switch, and only $30$ end with “No,” bounding how many can switch

Solution:

Assume $100$ students. The number of “Yes” answers rises from $50$ to $70,$ so at least $70-50=20$ students switched from “No” to “Yes”; thus $x\ge20.$

Since only $30$ students answer “No” at the end, at least $50-30=20$ of the original “Yes” students still answer “Yes,” so at most $80$ students switched; thus $x\le80.$

Both extremes are achievable, so the difference is $80-20=60.$

Thus, the correct answer is D.

7.

Shelby drives her scooter at a speed of $30$ miles per hour if it is not raining, and $20$ miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of $16$ miles in $40$ minutes. How many minutes did she drive in the rain?

$18$

$21$

$24$

$27$

$30$

Concepts:distance rate and time linear equation

Difficulty rating: 1240

Small Hint:

Let $t$ be the minutes in the rain; convert each speed to miles per minute

Big Hint:

Rain distance $20\cdot\dfrac{t}{60}$ plus sun distance $30\cdot\dfrac{40-t}{60}$ equals $16$

Solution:

Let $t$ be the number of minutes driven in the rain. She covers $20\cdot\frac{t}{60}$ miles in the rain and $30\cdot\frac{40-t}{60}$ miles in the sun.

Setting the total to $16$ gives $\frac{20t+30(40-t)}{60}=16,$ so $1200-10t=960$ and $t=24.$

Thus, the correct answer is C.

8.

Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea’s score was the median among all students, and hers was the highest score on her team. Andrea’s teammates Beth and Carla placed $37$ th and $64$ th, respectively. How many schools are in the city?

$22$

$23$

$24$

$25$

$26$

Concepts:median (data)parity inequality

Difficulty rating: 1490

Small Hint:

With $n$ schools there are $3n$ students, and the median sits at position $\dfrac{3n+1}{2}$

Big Hint:

$3n\ge64,$ and Andrea placed ahead of Beth’s $37$ th, so $\dfrac{3n+1}{2}\lt37$

Solution:

With $n$ schools there are $3n$ students. Carla placed $64$ th, so $3n\ge64$ and $n\ge22.$

The scores are distinct and Andrea is the median, so $3n$ is odd, forcing $n$ odd and $n\ge23.$

Andrea’s position is $\dfrac{3n+1}{2},$ and she beat Beth ( $37$ th), so $\dfrac{3n+1}{2}\lt37,$ giving $3n\lt73$ and $n\le24.$ The only odd value is $n=23.$

Thus, the correct answer is B.

9.

Let $n$ be the smallest positive integer such that $n$ is divisible by $20,$ $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n?$

$3$

$4$

$5$

$6$

$7$

Concepts:prime factorization perfect power least common multiple

Difficulty rating: 1520

Small Hint:

Write $n=2^a\cdot5^b;$ any extra prime factor only makes $n$ larger

Big Hint:

$n^2$ a cube forces $a$ and $b$ to be multiples of $3;$ $n^3$ a square forces $a$ and $b$ to be multiples of $2$

Solution:

To be smallest, $n$ uses only the primes of $20,$ so $n=2^a\cdot5^b$ with $a\ge2$ and $b\ge1.$

Since $n^2=2^{2a}5^{2b}$ is a perfect cube, $a$ and $b$ are multiples of $3.$ Since $n^3=2^{3a}5^{3b}$ is a perfect square, $a$ and $b$ are multiples of $2.$ Hence $a$ and $b$ are multiples of $6.$

The smallest choice is $a=b=6,$ so $n=2^6\cdot5^6=10^6=1{,}000{,}000,$ which has $7$ digits.

Thus, the correct answer is E.

10.

The average of the numbers $1,$ $2,$ $3,$ $\ldots,$ $98,$ $99,$ and $x$ is $100x.$ What is $x?$

$\dfrac{49}{101}$

$\dfrac{50}{101}$

$\dfrac{1}{2}$

$\dfrac{51}{101}$

$\dfrac{50}{99}$

Concepts:mean summation linear equation

Difficulty rating: 1410

Small Hint:

The sum $1+2+\cdots+99$ equals $4950$

Big Hint:

$\dfrac{4950+x}{100}=100x$

Solution:

The numbers $1$ through $99$ sum to $\dfrac{99\cdot100}{2}=4950.$

The average condition is $\frac{4950+x}{100}=100x,$ so $4950+x=10000x$ and $9999x=4950.$

Thus $x=\dfrac{4950}{9999}=\dfrac{50}{101}.$

Thus, the correct answer is B.

11.

A palindrome between $1000$ and $10{,}000$ is chosen at random. What is the probability that it is divisible by $7?$

$\dfrac{1}{10}$

$\dfrac{1}{9}$

$\dfrac{1}{7}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

Concepts:palindrome divisibility basic probability

Difficulty rating: 1500

Small Hint:

A four-digit palindrome $\overline{abba}$ equals $1001a+110b$

Big Hint:

$1001=7\cdot11\cdot13$ is divisible by $7,$ but $110$ is not

Solution:

A four-digit palindrome has the form $\overline{abba}=1001a+110b$ with $1\le a\le9$ and $0\le b\le9.$

Since $1001=7\cdot11\cdot13$ is divisible by $7$ and $110$ is not, the number is divisible by $7$ exactly when $b$ is divisible by $7,$ that is $b=0$ or $b=7.$

For each $a,$ that is $2$ of the $10$ choices of $b,$ a probability of $\dfrac{2}{10}=\dfrac15.$

Thus, the correct answer is E.

12.

For what value of $x$ does $\begin{aligned} &\log_{\sqrt2}\sqrt{x}+\log_2 x \\ &\quad {}+\log_4\left(x^2\right)+\log_8\left(x^3\right) \\ &\quad {}+\log_{16}\left(x^4\right)=40? \end{aligned}$

$8$

$16$

$32$

$256$

$1024$

Concepts:logarithm

Difficulty rating: 1500

Small Hint:

Convert every term to base $2;$ let $L=\log_2 x$

Big Hint:

Each of the five terms simplifies to $L$

Solution:

Let $L=\log_2 x.$ Converting each term to base $2:$

$\log_{\sqrt2}\sqrt{x}=\dfrac{\frac{L}{2}}{\frac{1}{2}}=L,$ $\log_2 x=L,$ $\log_4 x^2=\dfrac{2L}{2}=L,$ $\log_8 x^3=\dfrac{3L}{3}=L,$ and $\log_{16}x^4=\dfrac{4L}{4}=L.$

The equation becomes $5L=40,$ so $L=8$ and $x=2^8=256.$

Thus, the correct answer is D.

13.

In $\triangle ABC,$ $\cos(2A-B)+\sin(A+B)=2$ and $AB=4.$ What is $BC?$

$\sqrt{2}$

$\sqrt{3}$

$2$

$2\sqrt{2}$

$2\sqrt{3}$

Concepts:trigonometry special right triangle bounding to limit cases

Difficulty rating: 1560

Small Hint:

Both a cosine and a sine are at most $1,$ so each must equal $1$

Big Hint:

Solve $2A-B=0^\circ$ and $A+B=90^\circ$ to identify the triangle

Solution:

A cosine plus a sine equals $2$ only when each equals $1.$ So $\cos(2A-B)=1$ and $\sin(A+B)=1,$ giving $2A-B=0^\circ$ and $A+B=90^\circ.$

Solving, $A=30^\circ$ and $B=60^\circ,$ so $\triangle ABC$ is a $30\text{-}60\text{-}90$ right triangle with the right angle at $C.$

With hypotenuse $AB=4,$ the side $BC$ opposite the $30^\circ$ angle is half the hypotenuse, so $BC=2.$

Thus, the correct answer is C.

14.

Let $a,$ $b,$ $c,$ $d,$ and $e$ be positive integers with $a+b+c+d+e=2010,$ and let $M$ be the largest of the sums $a+b,$ $b+c,$ $c+d,$ and $d+e.$ What is the smallest possible value of $M?$

$670$

$671$

$802$

$803$

$804$

Concepts:optimization inequality extremal argument

Difficulty rating: 1670

Small Hint:

$(a+b)+c+(d+e)=2010,$ and each of $a+b,$ $c,$ and $d+e$ is at most $M$

Big Hint:

This forces $3M\ge2010;$ rule out $M=670,$ then build an example reaching the bound

Solution:

Each of $a+b,$ $d+e,$ and $c$ is at most $M$ (note $c\le c+d\le M$ ). Adding, $2010=(a+b)+c+(d+e)$ $\le3M,$ so $M\ge670.$

If $M=670,$ then $c=670,$ but then $b+c\ge671\gt M,$ a contradiction. Hence $M\ge671.$

The value $671$ is reached by $(a,b,c,d,e)=$ $(669,1,670,1,669),$ whose consecutive-pair sums are $670,671,671,670.$

Thus, the correct answer is B.

15.

For how many ordered triples $(x, y, z)$ of nonnegative integers less than $20$ are there exactly two distinct elements in the set $\{i^x, (1+i)^y, z\},$ where $i=\sqrt{-1}?$

$149$

$205$

$215$

$225$

$235$

Concepts:complex number casework basic counting

Difficulty rating: 2070

Small Hint:

$|i^x|=1$ always, while $(1+i)^y$ has magnitude $2^{\frac{y}{2}}$

Big Hint:

Split into three cases by which two of the three entries coincide, keeping the third distinct

Solution:

We need exactly two of $i^x,$ $(1+i)^y,$ $z$ equal, with the third different. The three cases are the three possible equal pairs.

Case $i^x=(1+i)^y:$ since $|i^x|=1$ but $|(1+i)^y|=2^{\frac{y}{2}}\gt1$ for $y\ge1,$ we need $y=0,$ so $(1+i)^0=1$ and $i^x=1,$ i.e. $x\in\{0,4,8,12,16\}.$ Then $z$ is any of the $19$ values other than $1.$ This gives $5\cdot19=95$ triples.

Case $i^x=z:$ the only nonnegative-integer value of $i^x$ is $1$ (with $x$ a multiple of $4$ ), so $z=1$ and $(1+i)^y\ne1,$ meaning $y\ge1.$ This gives $5\cdot19=95$ triples.

Case $(1+i)^y=z:$ since $(1+i)^2=2i,$ the power $(1+i)^y$ is a nonnegative integer below $20$ only for $y=0$ (value $1$ ) or $y=8$ (value $16$ ). If $y=0,\ z=1,$ we need $i^x\ne1,$ so $x$ is not a multiple of $4$ ( $15$ values). If $y=8,\ z=16,$ then $i^x$ is never $16,$ so $x$ is free ( $20$ values). This gives $15+20=35$ triples.

Altogether $95+95+35=225.$

Thus, the correct answer is D.

16.

Positive integers $a,$ $b,$ and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3, \ldots, 2010\}.$ What is the probability that $abc+ab+a$ is divisible by $3?$

$\dfrac{1}{3}$

$\dfrac{29}{81}$

$\dfrac{31}{81}$

$\dfrac{11}{27}$

$\dfrac{13}{27}$

Concepts:modular arithmetic basic probability casework

Difficulty rating: 1810

Small Hint:

Factor $abc+ab+a=a(bc+b+1);$ each residue mod $3$ is equally likely

Big Hint:

If $a$ is divisible by $3$ it works; otherwise require $bc+b+1$ to be divisible by $3$

Solution:

Factor $abc+ab+a=a(bc+b+1).$ Since $2010$ is a multiple of $3,$ each of $a, b, c$ is uniform modulo $3.$

If $a$ is divisible by $3$ (probability $\tfrac13$ ), the product is divisible by $3.$

If $a$ is not divisible by $3$ (probability $\tfrac23$ ), we need $bc+b+1$ to be divisible by $3.$ Checking residues, this holds exactly when $(b,c)\equiv(1,1)$ or $(2,0)\pmod3,$ a probability of $\tfrac13\cdot\tfrac13+\tfrac13\cdot\tfrac13=\tfrac29.$

The total probability is $\frac13+\frac23\cdot\frac29=\frac13+\frac{4}{27}=\frac{13}{27}.$

Thus, the correct answer is E.

17.

The entries in a $3\times3$ array include all the digits from $1$ through $9,$ arranged so that the entries in every row and column are in increasing order. How many such arrays are there?

$18$

$24$

$36$

$42$

$60$

Concepts:arrangements with restrictions casework

Difficulty rating: 1980

Small Hint:

The top-left entry is forced to be $1$ and the bottom-right to be $9$

Big Hint:

Split into cases by the center entry, which must be $4, 5,$ or $6$

Solution:

Write $a_{ij}$ for the entry in row $i,$ column $j.$ The conditions force $a_{11}=1,$ $a_{33}=9,$ and $a_{22}\in\{4,5,6\}.$

If $a_{22}=4,$ then $\{a_{12},a_{21}\}=\{2,3\}$ and $\{5,6,7,8\}$ split as complementary pairs filling the rest of the last row and column: $\binom42=6$ splits times $2$ orders for $\{2,3\}$ gives $12$ arrays. By symmetry $a_{22}=6$ also gives $12.$

If $a_{22}=5,$ then $\{a_{12},a_{13},a_{23}\}$ and $\{a_{21},a_{31},a_{32}\}$ are complementary subsets of $\{2,3,4,6,7,8\}$ subject to the ordering constraints. The first set can be any three-element subset except $\{2,3,4\}$ or $\{6,7,8\},$ giving $\binom63-2=18$ arrays.

Altogether $12+12+18=42.$

Thus, the correct answer is D.

18.

A frog makes $3$ jumps, each exactly $1$ meter long. The directions of the jumps are chosen independently and at random. What is the probability that the frog’s final position is no more than $1$ meter from its starting position?

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Concepts:geometric probability random walk

Difficulty rating: 2030

Small Hint:

Anchor the middle jump on a fixed segment; the start and end depend on two independent angles

Big Hint:

Reduce the condition to a region in the plane of those two angles, then compare areas

Solution:

This is a continuous (geometric) probability. Anchor the second jump from $P=(0,0)$ to $Q=(1,0),$ and let $\alpha,\beta$ be the directions of the first and third jumps, so the start is $A=(\cos\alpha,\sin\alpha)$ and the end is $B=(1+\cos\beta,\sin\beta).$

Taking $0\le\alpha\le\pi$ and $0\le\beta\le2\pi,$ the requirement $AB\le1$ holds exactly when $\alpha\le\beta\le\pi.$

In the $\alpha\beta$ -rectangle of area $2\pi^2,$ the favorable region is a triangle of area $\tfrac{\pi^2}{2},$ so the probability is $\dfrac{\frac{\pi^2}{2}}{2\pi^2}=\dfrac14.$

Thus, the correct answer is C.

19.

A high school basketball game between the Raiders and the Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $100$ points. What was the total number of points scored by the two teams in the first half?

$30$

$31$

$32$

$33$

$34$

Concepts:geometric sequence arithmetic sequence bounding to limit cases

Difficulty rating: 2180

Small Hint:

Let the Raiders score $a, ar, ar^2, ar^3$ and the Wildcats $a, a+d, a+2d, a+3d,$ tied at $a$

Big Hint:

The totals differ by $1$ and both stay under $100,$ which forces small values

Solution:

Let the Raiders score $a, ar, ar^2, ar^3$ (increasing geometric, $r\gt1$ ) and the Wildcats $a, a+d, a+2d, a+3d$ (increasing arithmetic), tied in the first quarter at $a.$

Write $r=\frac{m}{n}$ in lowest terms, with $m\gt n.$ Since $ar^3$ is an integer, $a$ is divisible by $n^3;$ put $a=A n^3.$ The Raiders’ total is $R=A(n^3+mn^2+m^2n+m^3).$ Since $m^3\lt R\le100,$ we have $m\le4.$ The pair $(4,3)$ already makes the parenthesized sum $175,$ so the only possible coprime pairs $(m,n)$ are $(2,1),(3,1),(3,2),(4,1).$

The corresponding base values of $\frac{R}{A}$ are $15,40,65,85.$ For $(4,1)$ and $(3,2),$ the bound forces $A=1,$ but $R-1=4a+6d$ gives a nonintegral $d.$ For $(3,1),$ it would give $36A-1=6d,$ which is impossible modulo $6.$

For $(m,n)=(2,1),$ we have $R=15A$ and $a=A,$ so $11A-1=6d.$ Thus $A\equiv5\pmod6,$ and $R\le100$ forces $A=5.$ Then $d=9,$ giving Raiders scores $5,10,20,40$ and Wildcats scores $5,14,23,32.$ The Raiders win $75$ to $74.$

The first-half total is $(5+10)+(5+14)=34.$

Thus, the correct answer is E.

20.

A geometric sequence $(a_n)$ has $a_1=\sin x,$ $a_2=\cos x,$ and $a_3=\tan x$ for some real number $x.$ For what value of $n$ does $a_n=1+\cos x?$

$4$

$5$

$6$

$7$

$8$

Concepts:geometric sequence trigonometric identity

Difficulty rating: 2240

Small Hint:

The common ratio is $r=\dfrac{\cos x}{\sin x}=\cot x;$ compute $a_4=a_3\cdot r$

Big Hint:

Show $a_4=1$ and $1+\cos x=r^4,$ then write $1+\cos x$ as $a_4\cdot r^4$

Solution:

The common ratio is $r=\dfrac{a_2}{a_1}=\cot x.$ Then $a_4=a_3\cdot r=\tan x\cot x=1.$

From $a_3=a_1r^2,$ we get $\tan x=\sin x\cot^2 x=\dfrac{\cos^2 x}{\sin x},$ so $\sin^2 x=\cos^3 x,$ i.e. $(\cos^2 x)(1+\cos x)=1.$

Hence $1+\cos x=\dfrac{1}{\cos^2 x}.$ Also $r^2=\dfrac{\cos^2 x}{\sin^2 x}=\dfrac{\cos^2 x}{\cos^3 x}=\dfrac{1}{\cos x},$ so $r^4=\dfrac{1}{\cos^2 x}=1+\cos x.$

Therefore $1+\cos x=a_4\cdot r^4=a_8,$ so $n=8.$

Thus, the correct answer is E.

21.

Let $a\gt0,$ and let $P(x)$ be a polynomial with integer coefficients such that $\begin{aligned} &P(1)=P(3)=P(5) \\ &\quad {}=P(7)=a, \end{aligned}$ and $\begin{aligned} &P(2)=P(4)=P(6) \\ &\quad {}=P(8)=-a. \end{aligned}$ What is the smallest possible value of $a?$

$105$

$315$

$945$

$7!$

$8!$

Concepts:polynomial divisibility least common multiple

Difficulty rating: 2300

Small Hint:

$P(x)-a=(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ for some integer polynomial $Q$

Big Hint:

Evaluate at $x=2,4,6,8$ to force $2a$ to be divisible by $\operatorname{lcm}(15,9,105)$

Solution:

Since $1, 3, 5, 7$ are roots of $P(x)-a,$ write $P(x)-a=(x-1)(x-3)$ $\cdot(x-5)(x-7)Q(x)$ with $Q$ having integer coefficients.

Evaluating at $x=2, 4, 6, 8$ (where $P=-a$ ) gives $\begin{aligned} -2a &=-15\,Q(2) \\ &=9\,Q(4) \\ &=-15\,Q(6) \\ &=105\,Q(8). \end{aligned}$

So $15, 9,$ and $105$ all divide $2a,$ hence $\operatorname{lcm}(15,9,105)=315$ divides $2a.$ Since $315$ is odd, $a$ is divisible by $315,$ so $a\ge315.$

To attain the bound, let $\begin{aligned} F(x)&=(x-1)(x-3) \\ &\quad\cdot(x-5)(x-7), \\ Q(x)&=42+(x-2)(x-6) \\ &\quad\cdot(60-8x), \\ P(x)&=315+F(x)Q(x). \end{aligned}$ This integer polynomial equals $315$ at $1,3,5,7.$ At $2,4,$ the pairs $(F(x),Q(x))$ are $(-15,42),(9,-70),$ and at $6,8$ they are $(-15,42),(105,-6).$ Thus $F(x)Q(x)=-630$ each time and $P(x)=-315.$ Hence the bound is attainable.

Thus, the correct answer is B.

22.

Let $ABCD$ be a cyclic quadrilateral. The side lengths of $ABCD$ are distinct integers less than $15$ such that $BC\cdot CD=AB\cdot DA.$ What is the largest possible value of $BD?$

$\sqrt{\dfrac{325}{2}}$

$\sqrt{185}$

$\sqrt{\dfrac{389}{2}}$

$\sqrt{\dfrac{425}{2}}$

$\sqrt{\dfrac{533}{2}}$

Concepts:cyclic quadrilateral Ptolemy’s Theorem optimization

Difficulty rating: 2420

Small Hint:

Let $a=AB,$ $b=BC,$ $c=CD,$ $d=DA$ with $bc=ad=k;$ comparing areas gives $(ab+cd)\cdot AC=2k\cdot BD$

Big Hint:

With Ptolemy’s $AC\cdot BD=ac+bd,$ eliminate $AC$ to get $BD^2=\tfrac12(a^2+b^2+c^2+d^2)$

Solution:

Let $a=AB,$ $b=BC,$ $c=CD,$ $d=DA$ and $k=bc=ad.$ Writing each triangle’s area in terms of the circumradius and using $[ABC]+[CDA]=$ $[BCD]+[ABD]$ gives $(ab+cd)\cdot AC=2k\cdot BD.$

Ptolemy’s theorem gives $AC\cdot BD=ac+bd.$ Eliminating $AC,$ $\begin{aligned} BD^2 &=\frac{(ac+bd)(ab+cd)}{2k} \\ &=\frac12\left(a^2+b^2+c^2+d^2\right). \end{aligned}$

The sides are distinct integers below $15$ with $bc=ad,$ so neither $11$ nor $13$ can appear (each is prime and would need a matching factor on the other side).

If the largest side is at most $12,$ the four squares sum to at most $12^2+10^2+9^2+8^2=389,$ because $11$ is unavailable.

Now suppose the largest side is $14,$ and write the others as $s_1\gt s_2\gt s_3.$ The product condition must pair $14$ with $s_3,$ so $14s_3=s_1s_2.$ Hence one of $s_1,s_2$ equals $7.$ If $s_1=7,$ the sum of squares is less than $14^2+7^2+6^2+5^2=306.$ If $s_2=7,$ then $s_1=2s_3,$ so the largest possibility is $(s_1,s_2,s_3)=(12,7,6).$ Thus $\begin{aligned} 2BD^2 &=14^2+12^2+7^2+6^2 \\ &=425, \end{aligned}$ so $BD\le\sqrt{\dfrac{425}{2}}.$ Equality is attained by the cyclic quadrilateral with side order $(a,b,c,d)=(14,12,7,6),$ for which $bc=ad=84.$

Thus, the correct answer is D.

23.

Monic quadratic polynomials $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23,$ $-21,$ $-17,$ and $-15,$ and $Q(P(x))$ has zeros at $x=-59,$ $-57,$ $-51,$ and $-49.$ What is the sum of the minimum values of $P(x)$ and $Q(x)?$

$-100$

$-82$

$-73$

$-64$

$0$

Concepts:polynomial completing the square symmetry (algebra)

Difficulty rating: 2420

Small Hint:

Write $P(x)=(x-h_1)^2-k_1^2$ and $Q(x)=(x-h_2)^2-k_2^2;$ the zeros of $P(Q(x))$ are symmetric about $x=h_2$

Big Hint:

The minimum values are $-k_1^2$ and $-k_2^2;$ find $k_1,k_2$ from the spacings of the given zeros

Solution:

If $P$ had only one real root, then $P(Q(x))=0$ would have at most two real solutions, not four. Thus $P$ has two distinct real roots, and the same argument applies to $Q.$ Write $P(x)=(x-h_1)^2-k_1^2$ and $Q(x)=(x-h_2)^2-k_2^2,$ with $k_1,k_2\gt0$ and minimum values $-k_1^2$ and $-k_2^2.$

The zeros of $P(Q(x))$ occur where $Q(x)=h_1\pm k_1;$ their four solutions are symmetric about $h_2,$ so $h_2$ is the average $\tfrac{-23-21-17-15}{4}=-19.$ Then $Q(-15)-Q(-17)$ $=(16-k_2^2)-(4-k_2^2)$ $=12,$ and this difference equals $2k_1,$ so $k_1=6.$

Symmetrically, $h_1=\tfrac{-59-57-51-49}{4}=-54,$ and $P(-49)-P(-51)$ $=(25-k_1^2)-(9-k_1^2)$ $=16=2k_2,$ so $k_2=8.$

The sum of the minimum values is $-k_1^2-k_2^2=-36-64=-100.$

Thus, the correct answer is A.

24.

The set of real numbers $x$ for which $\begin{aligned} &\frac{1}{x-2009}+\frac{1}{x-2010} \\ &\quad {}+\frac{1}{x-2011}\ge1 \end{aligned}$ is the union of intervals of the form $a\lt x\le b.$ What is the sum of the lengths of these intervals?

$\dfrac{1003}{335}$

$\dfrac{1004}{335}$

$3$

$\dfrac{403}{134}$

$\dfrac{202}{67}$

Concepts:rational equation inequality Vieta’s Formulas

Difficulty rating: 2320

Small Hint:

The left side is decreasing on each interval between its vertical asymptotes, so each solution interval ends at a root of the equation $=1$

Big Hint:

The three right endpoints are the roots of a cubic; sum them with Vieta

Solution:

Let $f(x)$ be the left-hand side. On each interval between consecutive asymptotes $2009, 2010, 2011,$ the function $f$ is decreasing, and $f\lt1$ for all $x\lt2009.$

On each of $(2009,2010),$ $(2010,2011),$ and $(2011,\infty),$ the solution is the part from the left asymptote up to a value $x_i$ where $f(x_i)=1.$ So the solution set consists of three intervals with left endpoints $2009, 2010, 2011$ and right endpoints $x_1, x_2, x_3.$

The total length is $(x_1-2009)$ $+(x_2-2010)$ $+(x_3-2011)$ $=x_1+x_2+x_3-6030.$

Clearing denominators in $f(x)=1$ gives $\begin{aligned} &x^3 \\ &\quad \small{}-(2009+2010+2011+3)x^2 \\ &\quad {}+\cdots=0, \end{aligned}$ whose roots are $x_1, x_2, x_3.$ By Vieta, $x_1+x_2+x_3=6033,$ so the sum of lengths is $6033-6030=3.$

Thus, the correct answer is C.

25.

For every integer $n\ge2,$ let $\operatorname{pow}(n)$ be the largest power of the largest prime that divides $n.$ For example, $\operatorname{pow}(144)=\operatorname{pow}(2^4\cdot3^2)=3^2.$ What is the largest integer $m$ such that $2010^m$ divides $\prod_{n=2}^{5300}\operatorname{pow}(n)?$

$74$

$75$

$76$

$77$

$78$

Concepts:prime factorization prime counting integers in a range

Difficulty rating: 2640

Small Hint:

$2010=2\cdot3\cdot5\cdot67;$ find the exponent of each prime in the product and take the minimum

Big Hint:

$\operatorname{pow}(n)$ contributes a prime $p$ only when $p$ is the largest prime factor of $n$

Solution:

Since $2010=2\cdot3\cdot5\cdot67,$ write the product as $2^A3^B5^C67^D$ times a factor coprime to all four primes; then $m=\min(A,B,C,D).$

Prime $2:$ $\operatorname{pow}(n)$ is a power of $2$ only when $n=2^k.$ Since $2^{12}=4096\lt5300\lt2^{13},$ the values $k=1,\ldots,12$ contribute $A=1+2+\cdots+12=78.$

Prime $67:$ $\operatorname{pow}(n)=67$ when $67$ is the largest prime factor, i.e. $n=67j$ with $1\le j\le79$ and every prime factor of $j$ at most $67;$ excluding $j=67, 71, 73, 79$ leaves $75$ values. The one $n$ with $\operatorname{pow}(n)=67^2$ is $n=67^2\lt5300,$ adding $2.$ So $D=75+2=77.$

Prime $3:$ For $n=2^a3^b$ with $b=1,2,3,4,$ the numbers of allowable exponents $a$ are $11,10,8,7,$ respectively. These terms alone contribute $\begin{aligned} B&\ge11+2\cdot10 \\ &\quad+3\cdot8+4\cdot7=83. \end{aligned}$

Prime $5:$ Write $n=2^a3^c5^b.$ For $b=1,$ the counts over $c=0,\ldots,6$ total $11+9+7+6+4+3+1=41.$ For $b=2,$ the counts total $8+7+5+3+2=25,$ each contributing two factors of $5.$ Hence $C\ge41+2\cdot25=91.$

Therefore $m=\min(78,B,C,77)=77.$

Thus, the correct answer is D.