2013 AMC 12B Solutions | LIVE by Po-Shen Loh

1.

On a particular January day, the high temperature in Lincoln, Nebraska, was $16$ degrees higher than the low temperature, and the average of the high and low temperatures was $3^\circ.$ In degrees, what was the low temperature in Lincoln that day?

$-13$

$-8$

$-5$

$-3$

$11$

Concepts:mean linear equation

Difficulty rating: 920

Small Hint:

The high and low differ by $16,$ so each is $8$ away from their average

Big Hint:

The low temperature is $8$ below the average of $3$

Solution:

The high exceeds the low by $16,$ so the low is $8$ below the average. Since the average is $3^\circ,$ the low temperature is $3 - 8 = -5^\circ.$ Thus, the correct answer is C.

2.

Mr. Green measures his rectangular garden by walking two of the sides and finds that it is $15$ steps by $20$ steps. Each of Mr. Green’s steps is $2$ feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden?

$600$

$800$

$1000$

$1200$

$1400$

Concepts:unit conversion area rate

Difficulty rating: 1020

Small Hint:

Convert steps to feet first: each step is $2$ feet

Big Hint:

Multiply the area in square feet by $\dfrac12$ pound per square foot

Solution:

The garden is $2\cdot 15 = 30$ feet by $2\cdot 20 = 40$ feet, an area of $1200$ square feet. At half a pound per square foot, Mr. Green expects $\tfrac12\cdot 1200 = 600$ pounds. Thus, the correct answer is A.

3.

When counting from $3$ to $201,$ $53$ is the $51$ st number counted. When counting backwards from $201$ to $3,$ $53$ is the $n$ th number counted. What is $n?$

$146$

$147$

$148$

$149$

$150$

Concepts:counting integers in a range

Difficulty rating: 1100

Small Hint:

Counting backwards, $201$ is the $1$ st number, $200$ is the $2$ nd, and so on

Big Hint:

The value $x$ is the $(202-x)$ th number counted

Solution:

Counting down from $201,$ the value $x$ is the $(202-x)$ th number. So $53$ is the $(202-53) = 149$ th number. Thus, the correct answer is D.

4.

Ray’s car averages $40$ miles per gallon of gasoline, and Tom’s car averages $10$ miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars’ combined rate of miles per gallon of gasoline?

$10$

$16$

$25$

$30$

$40$

Concepts:rate ratio and proportion

Difficulty rating: 1220

Small Hint:

Combined mileage is total miles divided by total gallons, not the average of the two rates

Big Hint:

If each drives $D$ miles, the gallons used are $\dfrac{D}{40}$ and $\dfrac{D}{10}$

Solution:

If each drives $D$ miles, together they cover $2D$ miles using $\dfrac{D}{40}+\dfrac{D}{10} = \dfrac{D}{8}$ gallons. The combined rate is $\dfrac{2D}{\frac{D}{8}} = 16$ miles per gallon. Thus, the correct answer is B.

5.

The average age of $33$ fifth-graders is $11.$ The average age of $55$ of their parents is $33.$ What is the average age of all of these parents and fifth-graders?

$22$

$23.25$

$24.75$

$26.25$

$28$

Concepts:mean

Difficulty rating: 1270

Small Hint:

The overall average is the total of all ages divided by the total number of people

Big Hint:

The total age is $55\cdot 33 + 33\cdot 11,$ spread over $55+33$ people

Solution:

The parents’ ages sum to $55\cdot 33$ and the fifth-graders’ to $33\cdot 11,$ a total of $33\cdot 66.$ Dividing by $88$ people gives $\dfrac{33\cdot 66}{88} = 24.75.$ Thus, the correct answer is C.

6.

Real numbers $x$ and $y$ satisfy the equation $x^2 + y^2 = 10x - 6y - 34.$ What is $x+y?$

$1$

$2$

$3$

$6$

$8$

Concepts:completing the square

Difficulty rating: 1370

Small Hint:

Move every term to one side and complete the square in both $x$ and $y$

Big Hint:

A sum of two squares equal to $0$ forces each square to be $0$

Solution:

Rearranging gives $x^2 - 10x + 25 + y^2 + 6y + 9$ $= 0,$ that is $(x-5)^2 + (y+3)^2 = 0.$ Hence $x = 5$ and $y = -3,$ so $x + y = 2.$ Thus, the correct answer is B.

7.

Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying “ $1$ ”, so Blair follows by saying “ $1,$ $2$ ”. Jo then says “ $1,$ $2,$ $3$ ”, and so on. What is the $53$ rd number said?

$2$

$3$

$5$

$6$

$8$

Concepts:triangular number

Difficulty rating: 1380

Small Hint:

The turn that ends at $n$ consists of someone saying $1, 2, \ldots, n$

Big Hint:

After the turn ending at $n,$ a total of $\tfrac12 n(n+1)$ numbers have been said

Solution:

After the turn that counts up to $n,$ exactly $1 + 2 + \cdots + n = \tfrac12 n(n+1)$ numbers have been said. For $n = 9$ that is $45.$ The next turn starts $1, 2, \ldots,$ so the $53$ rd number is the $8$ th number of that turn, namely $8.$ Thus, the correct answer is E.

8.

Line $\ell_1$ has equation $3x - 2y = 1$ and goes through $A = (-1, -2).$ Line $\ell_2$ has equation $y = 1$ and meets line $\ell_1$ at point $B.$ Line $\ell_3$ has positive slope, goes through point $A,$ and meets $\ell_2$ at point $C.$ The area of $\triangle ABC$ is $3.$ What is the slope of $\ell_3?$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$1$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

Concepts:coordinate geometry triangle area slope

Difficulty rating: 1460

Small Hint:

Find $B$ by solving $3x - 2y = 1$ together with $y = 1$

Big Hint:

The distance from $A$ to line $\ell_2$ is $3,$ so $\tfrac12\cdot BC\cdot 3 = 3$ gives $BC$

Solution:

Solving $3x - 2y = 1$ with $y = 1$ gives $B = (1, 1).$ The distance from $A = (-1, -2)$ to the line $y = 1$ is $3,$ so $\tfrac12\cdot BC\cdot 3 = 3$ gives $BC = 2.$ Then $C = (3, 1)$ or $C = (-1, 1);$ the latter makes $\ell_3$ vertical, so $C = (3, 1)$ and the slope is $\dfrac{1 - (-2)}{3 - (-1)} = \dfrac34.$ Thus, the correct answer is B.

9.

What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides $12!$ ?

$5$

$7$

$8$

$10$

$12$

Concepts:prime factorization factorial perfect square

Difficulty rating: 1510

Small Hint:

Factor $12! = 2^{10}\cdot 3^5\cdot 5^2\cdot 7\cdot 11$

Big Hint:

The largest square divisor keeps only even exponents; halving them gives the square root

Solution:

Since $12! = 2^{10}\cdot 3^5\cdot 5^2\cdot 7\cdot 11,$ the largest perfect square dividing it is $2^{10}\cdot 3^4\cdot 5^2,$ whose square root is $2^5\cdot 3^2\cdot 5.$ The exponents sum to $5 + 2 + 1 = 8.$ Thus, the correct answer is C.

10.

Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?

$62$

$82$

$83$

$102$

$103$

Concepts:system of equations invariant

Difficulty rating: 1550

Small Hint:

After $m$ exchanges at the first booth and $n$ at the second, Alex has $75-(2m-n)$ red and $75-(3n-m)$ blue tokens

Big Hint:

Exchanges stop when fewer than $2$ red and fewer than $3$ blue remain; the silver count is $m+n$

Solution:

After $m$ red-booth and $n$ blue-booth exchanges, Alex has $75 - (2m - n)$ red tokens, $75-(3n-m)$ blue tokens, and $m+n$ silver tokens. At termination the red count is $0$ or $1,$ and the blue count is $0,1,$ or $2.$ Solving the token equations over these six cases leaves only $(m,n)=(59,44),$ ending at $(1,2),$ or $(m,n)=(60,45),$ ending at $(0,0).$ The latter is unreachable: its final exchange would have to start at either $(-1,3)$ or $(2,-1).$ Hence Alex finishes with $59+44=103$ silver tokens. Thus, the correct answer is E.

11.

Two bees start at the same spot and fly at the same rate in the following directions. Bee $A$ travels $1$ foot north, then $1$ foot east, then $1$ foot upwards, and then continues to repeat this pattern. Bee $B$ travels $1$ foot south, then $1$ foot west, and then continues to repeat this pattern. In what directions are the bees traveling when they are exactly $10$ feet away from each other?

$A$ east, $B$ west

$A$ north, $B$ south

$A$ north, $B$ west

$A$ up, $B$ south

$A$ up, $B$ west

Concepts:3D geometry distance formula

Difficulty rating: 1610

Small Hint:

Use coordinates with east, north, up as $x, y, z;$ the bees only get farther apart over time

Big Hint:

Compute the distance right before and right after each foot to find where it crosses $10$

Solution:

Take east, north, up as $x, y, z.$ After $7$ feet bee $A$ is at $(2, 3, 2)$ and bee $B$ is at $(-3, -4, 0),$ a distance $\sqrt{78} \lt 10.$ On the next foot bee $A$ moves east to $(3, 3, 2)$ and bee $B$ moves west to $(-4, -4, 0),$ a distance $\sqrt{102} \gt 10.$ So they pass through $10$ feet apart while $A$ heads east and $B$ heads west. Thus, the correct answer is A.

12.

Cities $A,$ $B,$ $C,$ $D,$ and $E$ are connected by roads $AB,$ $AD,$ $AE,$ $BC,$ $BD,$ $CD,$ and $DE.$ How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit some cities more than once.)

$7$

$9$

$12$

$16$

$18$

Concepts:graph theory casework multiplication principle

Difficulty rating: 1670

Small Hint:

Cities $C$ and $E$ each touch only two roads, so they behave like detours on the $A$ – $D$ and $B$ – $D$ connections

Big Hint:

Count the trails from $A$ to $B$ on the reduced graph, then multiply by the independent detour choices

Solution:

City $E$ (roads $AE, DE$ ) is a detour on an $A$ – $D$ trip, and city $C$ (roads $BC, CD$ ) is a detour on a $B$ – $D$ trip. Replace them to get a graph on $A, B, D$ with two $A$ – $D$ connections, two $B$ – $D$ connections, and one $A$ – $B$ road. The trails from $A$ to $B$ using each once are of $4$ types: $ABDADB,$ $ADABDB,$ $ADBADB,$ and $ADBDAB.$ Each detour (through $E,$ through $C$ ) can be taken on either passage, so each type gives $4$ actual routes, for $4\cdot 4 = 16$ routes. Thus, the correct answer is D.

13.

The internal angles of quadrilateral $ABCD$ form an arithmetic progression. Triangles $ABD$ and $DCB$ are similar with $\angle DBA = \angle DCB$ and $\angle ADB = \angle CBD.$ Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the two largest angles of $ABCD?$

$210$

$220$

$230$

$240$

$250$

Concepts:arithmetic sequence angle chasing casework

Difficulty rating: 1700

Small Hint:

A triangle’s angles form an arithmetic progression exactly when its middle angle is $60^\circ$

Big Hint:

Let $\angle DBA = x$ and $\angle ADB = y;$ the four angles of $ABCD$ become $x, y, 180-y, 180-x,$ which must be an arithmetic progression with one triangle angle equal to $60^\circ$

Solution:

The angles of a triangle form an arithmetic progression exactly when the middle one is $60^\circ.$ With $\angle DBA = x$ and $\angle ADB = y,$ the four angles of $ABCD$ are $x, y, 180 - y, 180-x,$ which must itself be an arithmetic progression. In increasing order they are either $x,y,180-y,180-x$ or $x,180-y,y,180-x,$ giving $3y=x+180$ or $3y=360-x.$ One of the triangle angles $x,y,180-x-y$ is $60^\circ.$ Substitution leaves the angle sets $(60,80,100,120)$ and $(45,75,105,135).$ The two largest angles sum to at most $105 + 135 = 240.$ Thus, the correct answer is D.

14.

Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N.$ What is the smallest possible value of $N?$

$55$

$89$

$104$

$144$

$273$

Concepts:Fibonacci Diophantine Equation greatest common divisor

Difficulty rating: 1750

Small Hint:

In terms of the first two terms $a_1, a_2,$ the seventh term equals $5a_1 + 8a_2$

Big Hint:

Setting the two seventh terms equal gives $5(b_1 - a_1) = 8(a_2 - b_2);$ coprimality of $5$ and $8$ forces $b_1 - a_1$ to be a multiple of $8$

Solution:

A sequence starting $a_1, a_2$ has seventh term $5a_1 + 8a_2.$ For the two sequences, $5a_1 + 8a_2 = 5b_1 + 8b_2,$ so $5(b_1 - a_1) = 8(a_2 - b_2).$ Since $\gcd(5, 8) = 1,$ we need $b_1 - a_1$ to be a multiple of $8$ and $a_2 - b_2$ to be a multiple of $5.$ Taking $a_1 \lt b_1$ with nondecreasing terms gives $a_1 \le b_1 - 8 \le b_2 - 8 \le a_2 - 13.$ Choosing $a_1 = 0,$ $b_1 = b_2 = 8,$ $a_2 = 13$ yields $N = 5\cdot 0 + 8\cdot 13 = 104.$ Thus, the correct answer is C.

15.

The number $2013$ is expressed in the form

$2013 = \frac{a_1!\,a_2!\cdots a_m!}{b_1!\,b_2!\cdots b_n!},$

where $a_1 \ge a_2 \ge \cdots \ge a_m$ and $b_1 \ge b_2 \ge \cdots \ge b_n$ are positive integers and $a_1 + b_1$ is as small as possible. What is $|a_1 - b_1|\,?$

$1$

$2$

$3$

$4$

$5$

Concepts:prime factorization factorial bounding to limit cases

Difficulty rating: 1840

Small Hint:

Factor $2013 = 3\cdot 11\cdot 61;$ its largest prime is $61$

Big Hint:

Some $a_i!$ must reach $61,$ so $a_1 \ge 61;$ but $61!$ also contains the prime $59,$ which must be cancelled from the denominator

Solution:

Since $2013 = 3\cdot 11\cdot 61,$ the numerator needs a factorial at least $61!$ to supply the prime $61,$ so $a_1 \ge 61.$ But $61!$ also has a factor of $59,$ which $2013$ does not, so the denominator needs $b_1 \ge 59.$ Thus $a_1 + b_1 \ge 120,$ attained by $a_1 = 61,$ $b_1 = 59$ via $2013 = \dfrac{61!\,11!\,3!}{59!\,10!\,5!}.$ Then $|a_1 - b_1| = 2.$ Thus, the correct answer is B.

16.

Let $ABCDE$ be an equiangular convex pentagon of perimeter $1.$ The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s?$

$0$

$\dfrac{1}{2}$

$\dfrac{\sqrt5 - 1}{2}$

$\dfrac{\sqrt5 + 1}{2}$

$\sqrt5$

Concepts:equiangular polygon isosceles triangle invariant

Difficulty rating: 1890

Small Hint:

Each interior angle of an equiangular pentagon is $108^\circ,$ so every point of the star is an isosceles triangle with a $36^\circ$ apex

Big Hint:

Each star point’s two equal sides are the same fixed multiple $c$ of the pentagon side it sits on, so the star perimeter is $2c$ times the pentagon perimeter

Solution:

An equiangular pentagon has all interior angles $108^\circ,$ so each point of the star is an isosceles triangle with base angles $72^\circ$ and apex $36^\circ.$ By the equal base angles, each point contributes two sides that are the same fixed multiple $c$ of the pentagon side it rests on. Summing over the five points, the star perimeter equals $2c\cdot(\text{pentagon perimeter}) = 2c,$ independent of the individual side lengths. So $s$ is constant, and the difference between its maximum and minimum values is $0.$ Thus, the correct answer is A.

17.

Let $a,$ $b,$ and $c$ be real numbers such that

$a + b + c = 2$ and $a^2 + b^2 + c^2 = 12.$

What is the difference between the maximum and minimum possible values of $c?$

$2$

$\dfrac{10}{3}$

$4$

$\dfrac{16}{3}$

$\dfrac{20}{3}$

Concepts:quadratic inequality extremal argument

Difficulty rating: 1960

Small Hint:

Write $a + b = 2 - c$ and $a^2 + b^2 = 12 - c^2$

Big Hint:

Real $a, b$ with a given sum and sum of squares exist iff $(a+b)^2 \le 2(a^2 + b^2)$

Solution:

From the equations, $a + b = 2 - c$ and $a^2 + b^2 = 12 - c^2.$ Real numbers $a, b$ with a given sum and sum of squares exist iff $(a + b)^2 \le 2(a^2 + b^2),$ i.e. $(2 - c)^2 \le 2(12 - c^2).$ This simplifies to $(3c - 10)(c + 2) \le 0,$ so $-2 \le c \le \tfrac{10}{3}.$ The difference is $\tfrac{10}{3} - (-2) = \tfrac{16}{3}.$ Thus, the correct answer is D.

18.

Barbara and Jenna play the following game, in which they take turns. A number of coins lie on a table. When it is Barbara’s turn, she must remove $2$ or $4$ coins, unless only one coin remains, in which case she loses her turn. When it is Jenna’s turn, she must remove $1$ or $3$ coins. A coin flip determines who goes first. Whoever removes the last coin wins the game. Assume both players use their best strategy. Who will win when the game starts with $2013$ coins and when the game starts with $2014$ coins?

Barbara will win with $2013$ coins, and Jenna will win with $2014$ coins.

Jenna will win with $2013$ coins, and whoever goes first will win with $2014$ coins.

Barbara will win with $2013$ coins, and whoever goes second will win with $2014$ coins.

Jenna will win with $2013$ coins, and Barbara will win with $2014$ coins.

Whoever goes first will win with $2013$ coins, and whoever goes second will win with $2014$ coins.

Concepts:combinatorial game invariant modular arithmetic

Difficulty rating: 2070

Small Hint:

Track the number of coins modulo $5$

Big Hint:

Jenna can always restore a multiple of $5$ after her move, answering Barbara’s $2$ with $3$ and her $4$ with $1;$ decide who can maintain the invariant for each starting count

Solution:

Work modulo $5.$ With $2013 \equiv 3$ coins, Jenna wins either way: going first she takes $3$ to leave a multiple of $5,$ then answers Barbara’s $2$ with $3$ and $4$ with $1$ to keep multiples of $5,$ eventually taking the last coin; going second she keeps the count $\equiv 3 \pmod 5$ until Barbara is stuck at $3$ coins, must remove $2,$ and leaves Jenna the last coin. With $2014 \equiv 4$ coins, whoever goes first wins: Jenna first reduces to the $2013$ case, while Barbara first takes $4$ and then keeps multiples of $5.$ This is choice B. Thus, the correct answer is B.

19.

In triangle $ABC,$ $AB = 13,$ $BC = 14,$ and $CA = 15.$ Distinct points $D,$ $E,$ and $F$ lie on segments $BC,$ $CA,$ and $DE,$ respectively, such that $AD \perp BC,$ $DE \perp AC,$ and $AF \perp BF.$ The length of segment $DF$ can be written as $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m + n?$

$18$

$21$

$24$

$27$

$30$

Concepts:cyclic quadrilateral similarity altitude

Difficulty rating: 2140

Small Hint:

The altitude $AD$ in the $13$ - $14$ - $15$ triangle gives $BD = 5,$ $CD = 9,$ $AD = 12$

Big Hint:

Since $\angle AFB = \angle ADB = 90^\circ,$ points $A, B, D, F$ are concyclic; use similar right triangles to find $FE,$ then $DF = DE - FE$

Solution:

The altitude from $A$ to $BC$ gives $BD = 5,$ $CD = 9,$ $AD = 12.$ Because $DE \perp AC,$ triangle $AED \sim ADC,$ giving $DE = \tfrac{36}{5}$ and $AE = \tfrac{48}{5}.$ Since $\angle AFB = \angle ADB = 90^\circ,$ quadrilateral $ABDF$ is cyclic, so $\angle ABD = \angle AFE,$ making right triangles $ABD$ and $AFE$ similar: $\dfrac{FE}{5} = \dfrac{\frac{48}{5}}{12},$ so $FE = 4.$ Hence $DF = DE - FE$ $= \tfrac{36}{5} - 4$ $= \tfrac{16}{5},$ and $m + n = 21.$ Thus, the correct answer is B.

20.

For $135^\circ \lt x \lt 180^\circ,$ points $P = (\cos x, \cos^2 x),$ $Q = (\cot x, \cot^2 x),$ $R = (\sin x, \sin^2 x),$ and $S = (\tan x, \tan^2 x)$ are the vertices of a trapezoid. What is $\sin(2x)?$

$2 - 2\sqrt2$

$3\sqrt3 - 6$

$3\sqrt2 - 5$

$-\dfrac{3}{4}$

$1 - \sqrt3$

Concepts:parabola trigonometric identity trapezoid

Difficulty rating: 2270

Small Hint:

All four points lie on the parabola $y = t^2;$ the slope of the chord joining parameters $t_1$ and $t_2$ is $t_1 + t_2$

Big Hint:

On this range the parallel sides must be $QR$ and $PS,$ so $\cot x + \sin x = \tan x + \cos x$

Solution:

Each point $(t, t^2)$ lies on $y = t^2,$ and the chord through parameters $t_1, t_2$ has slope $t_1 + t_2.$ For $135^\circ \lt x \lt 180^\circ,$ both $\cos x$ and $\tan x$ lie between $\cot x$ and $\sin x,$ so $P$ and $S$ sit between $Q$ and $R$ and the parallel sides are $QR$ and $PS.$ Equal slopes give $\cot x + \sin x = \tan x + \cos x.$ Multiplying by $\sin x\cos x$ and simplifying yields $\cos x + \sin x - \sin x\cos x = 0.$ Squaring and using $2\sin x\cos x = \sin 2x$ gives $1 + \sin 2x = \tfrac14\sin^2 2x,$ whose only root in $(-1, 1)$ is $\sin 2x = 2 - 2\sqrt2.$ Thus, the correct answer is A.

21.

Consider the set of $30$ parabolas defined as follows: all parabolas have as focus the point $(0, 0)$ and the directrix lines have the form $y = ax + b$ with $a$ and $b$ integers such that $a \in \{-2, -1, 0, 1, 2\}$ and $b \in \{-3, -2, -1, 1, 2, 3\}.$ No three of these parabolas have a common point. How many points in the plane are on two of these parabolas?

$720$

$760$

$810$

$840$

$870$

Concepts:parabola complementary counting combinations

Difficulty rating: 2360

Small Hint:

Two parabolas sharing a focus meet in $2$ points unless their directrices are parallel with the focus not between them

Big Hint:

From all $\binom{30}{2}$ pairs, subtract the non-intersecting ones: equal slope and same-sign $y$ -intercepts

Solution:

Two parabolas with common focus $O$ meet in exactly $2$ points, except when their directrices are parallel and $O$ lies outside the strip between them, in which case they do not meet. The non-intersecting pairs have directrices of equal slope and $y$ -intercepts of the same sign. There are $5$ slopes, and for each, $2\binom{3}{2} = 6$ same-sign intercept pairs. Since every intersecting pair meets in $2$ points and no point lies on three parabolas, the total is $2\left(\binom{30}{2} - 5\cdot 6\right)$ $= 2(435 - 30)$ $= 810.$ Thus, the correct answer is C.

22.

Let $m \gt 1$ and $n \gt 1$ be integers. Suppose that the product of the solutions for $x$ of the equation

$\begin{aligned} &8(\log_n x)(\log_m x) - 7\log_n x \\ &\quad {}- 6\log_m x - 2013 = 0 \end{aligned}$

is the smallest possible integer. What is $m + n?$

$12$

$20$

$24$

$48$

$272$

Concepts:logarithm Vieta’s Formulas modular arithmetic

Difficulty rating: 2400

Small Hint:

Use $\log_n x = \dfrac{\log x}{\log n}$ to turn the equation into a quadratic in $\log x$

Big Hint:

The product of the roots satisfies $\log(x_1 x_2)$ $= \tfrac18(7\log m + 6\log n),$ so $(x_1 x_2)^8 = m^7 n^6;$ minimize this integer

Solution:

Writing $\log_n x = \tfrac{\log x}{\log n}$ and $\log_m x = \tfrac{\log x}{\log m},$ the equation becomes a quadratic in $\log x$ whose roots sum to $\log(x_1 x_2)$ $= \tfrac18(7\log m + 6\log n).$ Hence $N^8=m^7n^6,$ where $N=x_1x_2.$ For each prime dividing $mn,$ let its exponents in $m,n$ be $a,b.$ Then $7a+6b\equiv0\pmod8.$ An odd $a$ is impossible. If $a=0,$ then $b$ is a multiple of $4$ and this prime contributes at least $p^3,$ but some other prime must divide $m.$ If $a\equiv0\pmod8$ is positive, its contribution to $N$ is at least $p^7;$ if $a\equiv2\pmod8,$ then $b\equiv3\pmod4$ and the least contribution is $p^4.$ Every other positive even $a$ gives more. Thus the minimum uses only $p=2$ with $(a,b)=(2,3),$ giving $N=16,$ $m=4,$ and $n=8.$ So $m+n=12.$ Thus, the correct answer is A.

23.

Bernardo chooses a three-digit positive integer $N$ and writes both its base- $5$ and base- $6$ representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base- $10$ integers, he adds them to obtain an integer $S.$ For example, if $N = 749,$ Bernardo writes the numbers $10{,}444$ and $3{,}245,$ and LeRoy obtains the sum $S = 13{,}689.$ For how many choices of $N$ are the two rightmost digits of $S,$ in order, the same as those of $2N?$

$5$

$10$

$15$

$20$

$25$

Concepts:number base modular arithmetic casework

Difficulty rating: 2510

Small Hint:

Since $\mathrm{lcm}(5^2, 6^2, 10^2) = 900,$ the condition depends only on $N \bmod 900,$ so let $N$ range from $0$ to $899$

Big Hint:

Matching last digits forces the base- $5$ and base- $6$ units digits equal; working modulo $100$ pins down the allowed last-two-digit pairs

Solution:

Because $\mathrm{lcm}(25, 36, 100) = 900,$ the condition on $N$ depends only on $N \bmod 900,$ so consider $0 \le N \le 899.$ Let the last two base- $5$ digits be $a_1, a_0$ and the last two base- $6$ digits be $b_1,b_0.$ Modulo $10,$ the desired equality and $N\equiv a_0\pmod5$ force $a_0=b_0.$ The Chinese Remainder Theorem applied modulo $25$ and $36$ then gives $\begin{aligned} N&\equiv180a_1+150b_1 \\ &\quad {}+a_0\pmod{900}. \end{aligned}$ Comparing twice this residue modulo $100$ with the decimal number $10(a_1+b_1)+a_0+b_0$ reduces to $5a_1\equiv b_1\pmod{10}.$ Hence the valid pairs are exactly $(0,0),(2,0),(4,0),(1,5),(3,5).$ Each combines with $5$ choices of $a_0$ $(0 \le a_0 \le 4),$ giving $25$ values of $N.$ Thus, the correct answer is E.

24.

Let $ABC$ be a triangle where $M$ is the midpoint of $AC,$ and $CN$ is the angle bisector of $\angle ACB$ with $N$ on $AB.$ Let $X$ be the intersection of the median $BM$ and the bisector $CN.$ In addition $\triangle BXN$ is equilateral and $AC = 2.$ What is $BN^2?$

$\dfrac{10 - 6\sqrt2}{7}$

$\dfrac{2}{9}$

$\dfrac{5\sqrt2 - 3\sqrt3}{8}$

$\dfrac{\sqrt2}{6}$

$\dfrac{3\sqrt3 - 4}{5}$

Concepts:angle bisector equilateral triangle law of cosines similarity

Difficulty rating: 2600

Small Hint:

Let $\alpha = \angle ACN$ and $x = BN;$ the equilateral triangle gives $\angle BXC = \angle CNA = 120^\circ$

Big Hint:

Similar triangles yield $BC = \sqrt2$ and $CX = (\sqrt2 + 1)x;$ then apply the Law of Cosines in $\triangle BCX$

Solution:

Let $\alpha = \angle ACN = \angle NCB$ and $x = BN.$ Since $\triangle BXN$ is equilateral, $\angle BXC = \angle CNA = 120^\circ,$ which gives $\triangle ABC \sim \triangle BMC$ and $\triangle ANC \sim \triangle BXC.$ From the first, with $MC = \tfrac12 AC = 1,$ we get $\dfrac{BC}{2} = \dfrac{MC}{BC},$ so $BC = \sqrt2.$ From the second, $CX = (\sqrt2 + 1)x.$ The Law of Cosines in $\triangle BCX$ with $\angle BXC = 120^\circ$ gives $2 = x^2$ $+ (\sqrt2 + 1)^2 x^2$ $+ (\sqrt2 + 1)x^2$ $= (5 + 3\sqrt2)x^2.$ Hence $BN^2 = x^2$ $= \dfrac{2}{5 + 3\sqrt2}$ $= \dfrac{10 - 6\sqrt2}{7}.$ Thus, the correct answer is A.

25.

Let $G$ be the set of polynomials of the form

$\begin{aligned} &P(z) = z^n + c_{n-1}z^{n-1} + \cdots \\ &\quad {}+ c_2 z^2 + c_1 z + 50, \end{aligned}$

where $c_1,$ $c_2,$ $\ldots,$ $c_{n-1}$ are integers and $P(z)$ has $n$ distinct roots of the form $a + ib$ with $a$ and $b$ integers. How many polynomials are in $G?$

$288$

$528$

$576$

$992$

$1056$

Concepts:polynomial complex number factor counting casework

Difficulty rating: 2720

Small Hint:

Real coefficients pair nonreal roots as conjugates, so $P(z)$ factors into linear $(z - c)$ and quadratic $z^2 - 2az + (a^2 + b^2)$ pieces, each with constant term dividing $50$

Big Hint:

For each divisor $d$ of $50,$ count the basic factors of magnitude $d$ (solutions of $a^2 + b^2 = d,$ plus $z \pm d$ ), then multiply choices so the constant terms multiply to $50$

Solution:

Since the coefficients are real, nonreal roots occur in conjugate pairs, so $P(z)$ factors into distinct linear factors $(z - c)$ with $c \in \mathbb{Z}$ and quadratics $(z - (a+ib))(z - (a-ib))$ $= z^2 - 2az + (a^2 + b^2).$ Each factor’s constant term divides $50.$ For $d=1,2,5,10,25,50,$ the numbers of conjugate pairs with $a^2+b^2=d$ and $b\ne0$ are $1,2,4,4,5,6,$ respectively. Adding the two linear choices $z-d,z+d$ gives $|B_1|=3,$ $|B_2|=4,$ $|B_5|=6,$ $|B_{10}|=6,$ $|B_{25}|=7,$ and $|B_{50}|=8.$ The factor-magnitude partitions of $50$ using values greater than $1$ are $50,25\cdot2,10\cdot5,$ and $5\cdot5\cdot2.$ Distinct roots require choosing two different $B_5$ factors in the last case. Finally, account for the free presence of $z+1$ and $z^2+1$ (with $z-1$ forced by the sign of the remaining product), gives $\begin{aligned} &2^2\left(8 + 7\cdot 4 + 6\cdot 6 + 4\binom{6}{2}\right) \\ &\quad = 4(8 + 28 + 36 + 60) = 528. \end{aligned}$ Thus, the correct answer is B.