2014 AMC 12B Solutions | LIVE by Po-Shen Loh

1.

Leah has $13$ coins, all of which are pennies and nickels. If she had one more nickel than she has now, then she would have the same number of pennies and nickels. In cents, how much are Leah’s coins worth?

$33$

$35$

$37$

$39$

$41$

Concepts:linear equation money

Difficulty rating: 920

Small Hint:

Let $n$ be the number of nickels, so there are $13-n$ pennies

Big Hint:

Adding one nickel gives $n+1$ nickels, which must equal the $13-n$ pennies

Solution:

Let $n$ be the number of nickels, so Leah has $13-n$ pennies. One more nickel would give her $n+1$ nickels, and this equals the number of pennies: $n+1 = 13-n.$ Solving gives $n=6,$ so there are $6$ nickels and $7$ pennies.

The total value is $6\cdot5 + 7 = 37$ cents.

Thus, the correct answer is C.

2.

Orvin went to the store with just enough money to buy $30$ balloons. When he arrived he discovered that the store had a special sale on balloons: buy $1$ balloon at the regular price and get a second at $\tfrac13$ off the regular price. What is the greatest number of balloons Orvin could buy?

$33$

$34$

$36$

$38$

$39$

Concepts:ratio and proportion money

Difficulty rating: 1070

Small Hint:

Group the balloons into pairs, one at full price and one at $\tfrac23$ of the price

Big Hint:

Each pair costs $\tfrac53$ of a regular balloon’s price

Solution:

Under the sale, a pair of balloons costs $1 + \tfrac23 = \tfrac53$ times the regular price of one balloon.

Orvin’s money buys $30$ balloons at the regular price, so he can afford $30 \div \tfrac53 = 18$ pairs, which is $36$ balloons.

Thus, the correct answer is C.

3.

Randy drove the first third of his trip on a gravel road, the next $20$ miles on pavement, and the remaining one-fifth on a dirt road. In miles, how long was Randy’s trip?

$30$

$\dfrac{400}{11}$

$\dfrac{75}{2}$

$40$

$\dfrac{300}{7}$

Concepts:fraction linear equation

Difficulty rating: 1150

Small Hint:

The paved stretch is the fraction of the trip not on gravel or dirt

Big Hint:

That paved fraction $1-\tfrac13-\tfrac15$ equals $20$ miles

Solution:

The fraction of the trip on pavement is $1 - \tfrac13 - \tfrac15 = \tfrac{7}{15}.$

Since this equals $20$ miles, the whole trip is $20 \div \tfrac{7}{15} = \tfrac{300}{7}$ miles.

Thus, the correct answer is E.

4.

Susie pays for $4$ muffins and $3$ bananas. Calvin spends twice as much paying for $2$ muffins and $16$ bananas. A muffin is how many times as expensive as a banana?

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$\dfrac{7}{4}$

$2$

$\dfrac{13}{4}$

Concepts:linear equation ratio and proportion

Difficulty rating: 1230

Small Hint:

Let a muffin cost $m$ and a banana cost $b$

Big Hint:

Calvin spending twice as much means $2(4m+3b)=2m+16b$

Solution:

Let a muffin cost $m$ and a banana cost $b.$ Then $2(4m+3b) = 2m+16b.$

Expanding gives $8m+6b = 2m+16b,$ so $6m = 10b$ and $m = \tfrac53 b.$

Thus, the correct answer is B.

5.

Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2,$ and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window?

$26$

$28$

$30$

$32$

$34$

Concepts:ratio and proportion linear equation rectangle

Difficulty rating: 1400

Small Hint:

Let each pane be $2x$ wide and $5x$ tall

Big Hint:

The window’s width $4(2x)+5\cdot2$ equals its height $2(5x)+3\cdot2$

Solution:

Let each pane have width $2x$ and height $5x.$ The window is $4$ panes wide with $5$ vertical borders, so its width is $4(2x) + 5\cdot2 = 8x+10.$

It is $2$ panes tall with $3$ horizontal borders, so its height is $2(5x) + 3\cdot2 = 10x+6.$

Setting width equal to height gives $8x+10 = 10x+6,$ so $x=2$ and the side length is $10\cdot2+6 = 26.$

Thus, the correct answer is A.

6.

Ed and Ann both have lemonade with their lunch. Ed orders the regular size. Ann gets the large lemonade, which is $50\%$ more than the regular. After both consume $\tfrac34$ of their drinks, Ann gives Ed a third of what she has left, and $2$ additional ounces. When they finish their lemonades they realize that they both drank the same amount. How many ounces of lemonade did they drink together?

$30$

$32$

$36$

$40$

$50$

Concepts:fraction linear equation

Difficulty rating: 1460

Small Hint:

Let a regular lemonade be $a$ ounces, so the large is $\tfrac32 a$

Big Hint:

After drinking $\tfrac34,$ Ann has $\tfrac38 a$ left and hands Ed $\tfrac13$ of that plus $2$

Solution:

Let a regular lemonade hold $a$ ounces, so Ann’s large holds $\tfrac32 a.$ After each drinks $\tfrac34,$ Ann has $\tfrac14\cdot\tfrac32 a = \tfrac38 a$ left, and she gives Ed $\tfrac13\cdot\tfrac38 a + 2 = \tfrac18 a + 2$ ounces.

Ed drinks his full $a$ ounces plus that gift, and Ann drinks her $\tfrac32 a$ minus the gift. Setting these equal, $a + \left(\tfrac18 a + 2\right) = \tfrac32 a - \left(\tfrac18 a + 2\right),$ which gives $4 = \tfrac14 a,$ so $a = 16.$

Then Ed drank $16$ ounces and Ann drank $24$ ounces, for a total of $40$ ounces.

Thus, the correct answer is D.

7.

For how many positive integers $n$ is $\dfrac{n}{30-n}$ also a positive integer?

$4$

$5$

$6$

$7$

$8$

Concepts:divisibility factor algebraic manipulation

Difficulty rating: 1520

Small Hint:

Rewrite $\dfrac{n}{30-n} = \dfrac{30}{30-n} - 1$

Big Hint:

So $30-n$ must be a positive divisor of $30,$ and the quotient must be at least $2$

Solution:

Write $\dfrac{n}{30-n} = \dfrac{30}{30-n} - 1.$

For this to be a positive integer, $30-n$ must be a positive divisor of $30$ with $\dfrac{30}{30-n} \ge 2,$ i.e. $30-n \le 15.$

The divisors of $30$ that are at most $15$ are $1, 2, 3, 5, 6, 10, 15,$ giving $7$ values of $n$ (namely $15, 20, 24, 25, 27, 28, 29$ ).

Thus, the correct answer is D.

8.

In the addition shown below $A,$ $B,$ $C,$ and $D$ are distinct digits. How many different values are possible for $D?$

$\begin{array}{cccccc} & A & B & B & C & B \\ + & B & C & A & D & A \\ \hline & D & B & D & D & D \end{array}$

$2$

$4$

$7$

$8$

$9$

Concepts:cryptarithm place value casework

Difficulty rating: 1580

Small Hint:

The leftmost column gives $A+B=D$ with no carry, so $A+B \le 9$

Big Hint:

The second and fourth columns force $C=0$ and no carries anywhere

Solution:

The leftmost column shows $A+B = D$ with no carry out, so $A+B \le 9.$ Examining the tens and thousands columns (each of the form $C + \text{digit} + \text{carry}$ producing the same digit) forces $C = 0$ and eliminates all carries.

Every column then reduces to $A+B = D,$ with $A, B, C=0$ distinct. Since $A$ and $B$ are distinct positive digits, $D = A+B$ can be any value from $3$ up to $9,$ giving $7$ possibilities, for example $(A,B,C,D) = (1,2,0,3),$ $(1,3,0,4), \ldots,$ $(2,7,0,9).$

Thus, the correct answer is C.

9.

Convex quadrilateral $ABCD$ has $AB = 3,$ $BC = 4,$ $CD = 13,$ $AD = 12,$ and $\angle ABC = 90^\circ,$ as shown. What is the area of the quadrilateral?

$30$

$36$

$40$

$48$

$58.5$

Concepts:Pythagorean Theorem triangle area area decomposition

Difficulty rating: 1560

Small Hint:

Use right triangle $ABC$ to find diagonal $AC$

Big Hint:

Check whether $AC,$ $AD,$ $CD$ satisfy $AC^2 + AD^2 = CD^2$

Solution:

By the Pythagorean Theorem in right triangle $ABC,$ $AC = \sqrt{3^2+4^2} = 5.$

Since $5^2 + 12^2 = 13^2,$ the converse of the Pythagorean Theorem shows $\angle DAC = 90^\circ,$ so $\triangle DAC$ is right.

The area of $\triangle ABC$ is $\tfrac12\cdot3\cdot4 = 6$ and the area of $\triangle DAC$ is $\tfrac12\cdot5\cdot12 = 30.$ The quadrilateral has area $6 + 30 = 36.$

Thus, the correct answer is B.

10.

Danica drove her new car on a trip for a whole number of hours, averaging $55$ miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a $3$ -digit number with $a \ge 1$ and $a+b+c \le 7.$ At the end of the trip, the odometer showed $cba$ miles. What is $a^2 + b^2 + c^2?$

$26$

$27$

$36$

$37$

$41$

Concepts:place value divisibility least common multiple

Difficulty rating: 1680

Small Hint:

The distance driven is $cba - abc = 99(c-a)$

Big Hint:

It is also a multiple of $55,$ so it is a multiple of $\operatorname{lcm}(99,55)=495$

Solution:

The distance driven is $cba - abc = 99(c-a),$ a multiple of $9.$ Driving a whole number of hours at $55$ mph makes it a multiple of $55$ too, hence a multiple of $495.$

Since the odometer difference is at most a $3$ -digit number and $a \ge 1,$ the distance must be $495,$ so $c - a = 5.$

With $a \ge 1$ and $a+b+c \le 7,$ the only choice is $a=1,$ $c=6,$ $b=0.$ Then $a^2+b^2+c^2 = 1 + 0 + 36 = 37.$

Thus, the correct answer is D.

11.

A list of $11$ positive integers has a mean of $10,$ a median of $9,$ and a unique mode of $8.$ What is the largest possible value of an integer in the list?

$24$

$30$

$31$

$33$

$35$

Concepts:mean median (data)mode optimization

Difficulty rating: 1690

Small Hint:

The $11$ numbers sum to $110$

Big Hint:

Maximize one number by minimizing the other ten, keeping the $6$ th value $9$ and $8$ the unique mode

Solution:

The list sums to $11 \cdot 10 = 110.$ To maximize one entry, minimize the sum of the other ten.

Sorted, the sixth number must be $9$ (the median), and $8$ must appear more often than any other value. Trying $8$ three times, the smallest possible ten numbers are $1,1,8,8,8,9,9,10,10,11,$ which sum to $75$ and keep $8$ the unique mode.

The largest entry is then $110 - 75 = 35.$

Thus, the correct answer is E.

12.

A set $S$ consists of triangles whose sides have integer lengths less than $5,$ and no two elements of $S$ are congruent or similar. What is the largest number of elements that $S$ can have?

$8$

$9$

$10$

$11$

$12$

Concepts:triangle inequality similarity systematic listing

Difficulty rating: 1770

Small Hint:

List each triangle by its sides in nonincreasing order using values from $\{1,2,3,4\}$

Big Hint:

Drop those failing the triangle inequality, and count $2,2,1$ and $4,4,2$ only once since they are similar

Solution:

Write each triangle by its side lengths in nonincreasing order. Only one equilateral triangle is allowed (all are similar), and of the similar pair $2,2,1$ and $4,4,2$ only one may appear.

The remaining valid, pairwise non-similar triangles are $\begin{gathered} 4\,4\,3,\ 4\,4\,1,\ 4\,3\,3,\ 4\,3\,2,\ \\ 3\,3\,2,\ 3\,3\,1,\ 3\,2\,2, \end{gathered}$ seven in all. Together with one equilateral and one of the similar pair, $S$ has at most $9$ elements.

Thus, the correct answer is B.

13.

Real numbers $a$ and $b$ are chosen with $1 \lt a \lt b$ such that no triangle with positive area has side lengths $1,$ $a,$ and $b$ or $\tfrac1b,$ $\tfrac1a,$ and $1.$ What is the smallest possible value of $b?$

$\dfrac{3+\sqrt{3}}{2}$

$\dfrac{5}{2}$

$\dfrac{3+\sqrt{5}}{2}$

$\dfrac{3+\sqrt{6}}{2}$

$3$

Concepts:triangle inequality quadratic bounding to limit cases

Difficulty rating: 1870

Small Hint:

With $1 \lt a \lt b,$ no triangle $1, a, b$ means $b \ge a+1$

Big Hint:

No triangle $\tfrac1b, \tfrac1a, 1$ means $1 \ge \tfrac1a + \tfrac1b,$ i.e. $a \le \tfrac{b}{b-1}$

Solution:

Since $b$ is the largest of $1, a, b,$ no such triangle exists exactly when $b \ge a+1.$ Since $1$ is the largest of $\tfrac1b, \tfrac1a, 1,$ no such triangle exists exactly when $1 \ge \tfrac1a + \tfrac1b,$ that is $a \le \tfrac{b}{b-1}.$

Both conditions hold with $b$ smallest when $a+1 = b$ and $a = \tfrac{b}{b-1}$ meet, giving $b - 1 = \tfrac{b}{b-1},$ or $b^2 - 3b + 1 = 0.$

The root larger than $1$ is $b = \dfrac{3+\sqrt5}{2}.$

Thus, the correct answer is C.

14.

A rectangular box has a total surface area of $94$ square inches. The sum of the lengths of all its edges is $48$ inches. What is the sum of the lengths in inches of all of its interior diagonals?

$8\sqrt{3}$

$10\sqrt{2}$

$16\sqrt{3}$

$20\sqrt{2}$

$40\sqrt{2}$

Concepts:rectangular prism surface area algebraic manipulation

Difficulty rating: 1840

Small Hint:

Let the edges be $x, y, z,$ so $2(xy+yz+zx)=94$ and $4(x+y+z)=48$

Big Hint:

Use $(x+y+z)^2 =$ $x^2+y^2+z^2$ $+ 2(xy+yz+zx)$ to find each diagonal $\sqrt{x^2+y^2+z^2}$

Solution:

Let the edges be $x, y, z.$ Then $xy+yz+zx = 47$ and $x+y+z = 12.$ Therefore $\begin{gathered} x^2+y^2+z^2 = (x+y+z)^2 \\ {}- 2(xy+yz+zx) \\ = 144 - 94 \\ = 50. \end{gathered}$

Each of the $4$ interior diagonals has length $\sqrt{x^2+y^2+z^2} = \sqrt{50} = 5\sqrt2,$ so their total length is $4 \cdot 5\sqrt2 = 20\sqrt2.$

Thus, the correct answer is D.

15.

When $p = \sum_{k=1}^{6} k \ln k,$ the number $e^p$ is an integer. What is the largest power of $2$ that is a factor of $e^p?$

$2^{12}$

$2^{14}$

$2^{16}$

$2^{18}$

$2^{20}$

Concepts:logarithm prime factorization

Difficulty rating: 1950

Small Hint:

$k \ln k = \ln(k^k),$ so $e^p = \prod_{k=1}^{6} k^k$

Big Hint:

Only $2^2,$ $4^4,$ and $6^6$ contribute factors of $2$

Solution:

Since $k \ln k = \ln(k^k),$ the sum gives $p = \ln\left(\prod_{k=1}^{6} k^k\right),$ so $e^p = 1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4 \cdot 5^5 \cdot 6^6.$

The factors of $2$ come from $2^2$ (giving $2$ ), $4^4 = 2^8$ (giving $8$ ), and $6^6 = 2^6 \cdot 3^6$ (giving $6$ ). In total the exponent of $2$ is $2 + 8 + 6 = 16.$

Thus, the correct answer is C.

16.

Let $P$ be a cubic polynomial with $P(0) = k,$ $P(1) = 2k,$ and $P(-1) = 3k.$ What is $P(2) + P(-2)?$

$0$

$k$

$6k$

$7k$

$14k$

Concepts:polynomial system of equations symmetry (algebra)

Difficulty rating: 1950

Small Hint:

Write $P(x) = ax^3 + bx^2 + cx + k$ using $P(0)=k$

Big Hint:

Adding $P(1)$ and $P(-1)$ isolates $b,$ and $P(2)+P(-2) = 8b + 2k$

Solution:

Since $P(0) = k,$ write $P(x) = ax^3 + bx^2 + cx + k.$

Then $P(1) = a+b+c+k = 2k$ and $P(-1) = -a+b-c+k = 3k.$ Adding these gives $2b + 2k = 5k,$ so $2b = 3k.$

The odd-power terms cancel in the sum: $\begin{gathered} P(2)+P(-2) \\ = (8a+4b+2c+k) \\ {}+ (-8a+4b-2c+k) \\ = 8b + 2k. \end{gathered}$ Since $8b = 4(2b) = 12k,$ this equals $12k + 2k = 14k.$

Thus, the correct answer is E.

17.

Let $P$ be the parabola with equation $y = x^2$ and let $Q = (20, 14).$ There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r \lt m \lt s.$ What is $r + s?$

$1$

$26$

$40$

$52$

$80$

Concepts:parabola quadratic Vieta’s Formulas

Difficulty rating: 2010

Small Hint:

The line is $y = m(x-20) + 14;$ set it equal to $x^2$

Big Hint:

No intersection means the discriminant $m^2 - 80m + 56 \lt 0;$ use the sum of the roots $r, s$

Solution:

The line through $Q$ is $y = m(x-20) + 14.$ Substituting into $y = x^2$ gives $x^2 - mx + (20m - 14) = 0.$

There is no intersection exactly when this has no real root, i.e. when the discriminant $m^2 - 4(20m-14)$ $= m^2 - 80m + 56$ is negative. That happens between the two roots $r$ and $s$ of $m^2 - 80m + 56 = 0.$

By Vieta’s formulas, $r + s = 80.$

Thus, the correct answer is E.

18.

The numbers $1,$ $2,$ $3,$ $4,$ $5$ are to be arranged in a circle. An arrangement is bad if it is not true that for every $n$ from $1$ to $15$ one can find a subset of the numbers that appear consecutively on the circle that sum to $n.$ Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?

$1$

$2$

$3$

$4$

$5$

Concepts:circular arrangements complementary counting casework

Difficulty rating: 2150

Small Hint:

Sums $1$ through $5$ are always achievable with a single number

Big Hint:

If a consecutive block sums to $n,$ its complement sums to $15-n,$ so only $6$ and $7$ can fail

Solution:

Any single number covers sums $1$ through $5.$ If a consecutive block sums to $n,$ the remaining numbers form a consecutive block summing to $15 - n,$ so sums $10$ through $14$ are automatically covered as well. Thus an arrangement is bad only if it fails to produce $6$ or $7.$

If $6$ cannot be formed, rotate and reflect so the order is $1bc5e.$ The pair $\{b,c\}$ cannot be $\{2,3\}$ or $\{2,4\},$ so $e=2;$ avoiding the block $213$ then forces $14352.$ If $7$ cannot be formed, write the order as $2bc5e.$ Now $\{b,c\}$ cannot be $\{3,4\}$ or $\{1,4\},$ so $e=4,$ and avoiding $421$ forces $23154.$

These are the only two bad arrangements up to rotation and reflection.

Thus, the correct answer is B.

19.

A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?

$\dfrac{3}{2}$

$\dfrac{1+\sqrt{5}}{2}$

$\sqrt{3}$

$2$

$\dfrac{3+\sqrt{5}}{2}$

Concepts:cone sphere volume

Difficulty rating: 2220

Small Hint:

Take an axial cross-section, normalize the top radius to $1,$ and call the bottom and sphere radii $r$ and $a$

Big Hint:

Use the distance from the sphere center to a slanted side to relate $r$ and $a,$ then compare the frustum and sphere volumes

Solution:

Let the top radius be $1,$ the bottom radius $r,$ and the sphere radius $a.$ The sphere touches both bases, so the frustum height is $2a.$ In an axial cross-section put the sphere center at $(0,a)$ ; the right slanted side through $(r,0)$ and $(1,2a)$ has equation $2ax+(r-1)y-2ar=0.$ Its distance from $(0,a)$ is $a,$ so $(r+1)^2=4a^2+(r-1)^2,$ giving $r=a^2.$

The frustum volume is $\tfrac13 \pi (r^2 + r + 1)(2a).$ Setting it equal to twice the sphere volume $\tfrac43 \pi a^3$ and using $r = a^2$ yields $a^4 - 3a^2 + 1 = 0,$ that is $r^2 - 3r + 1 = 0.$

The positive root is $r = \dfrac{3+\sqrt5}{2}.$

Thus, the correct answer is E.

20.

For how many positive integers $x$ is $\log_{10}(x - 40)$ $+ \log_{10}(60 - x) \lt 2?$

$10$

$18$

$19$

$20$

infinitely many

Concepts:logarithm inequality quadratic

Difficulty rating: 2110

Small Hint:

Both logarithms require $40 \lt x \lt 60$

Big Hint:

Combining gives $(x-40)(60-x) \lt 100,$ which is $(x-50)^2 \gt 0$

Solution:

The logarithms are defined only when $x - 40 \gt 0$ and $60 - x \gt 0,$ so $40 \lt x \lt 60.$

Within this range the inequality becomes $(x-40)(60-x) \lt 100,$ which expands to $x^2 - 100x + 2500 \gt 0,$ i.e. $(x-50)^2 \gt 0.$ This holds for every $x \ne 50.$

The integers strictly between $40$ and $60$ except $50$ are $41, \ldots, 49$ and $51, \ldots, 59,$ which is $18$ values.

Thus, the correct answer is B.

21.

In the figure, $ABCD$ is a square of side length $1.$ The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE?$

$\dfrac12(\sqrt{6} - 2)$

$\dfrac14$

$2 - \sqrt{3}$

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

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

Concepts:right triangle trigonometry angle chasing

Difficulty rating: 2350

Small Hint:

Let $\theta = \angle DHG;$ in right triangle $GDH,$ $BE = \dfrac{1 - \cos\theta}{\sin\theta}$

Big Hint:

Along the top side, $CD = CF + FH + HD = 1$ simplifies to $2\sin\theta = 1$

Solution:

Let $x = BE = GH = CF$ and $\theta = \angle DHG = \angle AGJ$ $= \angle FKH,$ with $AD = GJ = HK = 1.$ In right triangle $GDH,$ $x \sin\theta = DG = 1 - \cos\theta,$ so $x = \dfrac{1 - \cos\theta}{\sin\theta}.$

Along side $CD,$ $\begin{gathered} 1 = CF + FH + HD \\ = x + \sin\theta + x\cos\theta. \end{gathered}$ Substituting for $x$ gives $\begin{gathered} 1 = \dfrac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta} \\ {}+ \sin\theta \\ = \dfrac{\sin^2\theta}{\sin\theta} \\ {}+ \sin\theta \\ = 2\sin\theta. \end{gathered}$

Hence $\sin\theta = \tfrac12,$ so $\theta = 30^\circ$ and $x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3.$

Thus, the correct answer is C.

22.

In a small pond there are eleven lily pads in a row labeled $0$ through $10.$ A frog is sitting on pad $1.$ When the frog is on pad $N,$ $0 \lt N \lt 10,$ it will jump to pad $N - 1$ with probability $\dfrac{N}{10}$ and to pad $N + 1$ with probability $1 - \dfrac{N}{10}.$ Each jump is independent of the previous jumps. If the frog reaches pad $0$ it will be eaten by a patiently waiting snake. If the frog reaches pad $10$ it will exit the pond, never to return. What is the probability that the frog will escape being eaten by the snake?

$\dfrac{32}{79}$

$\dfrac{161}{384}$

$\dfrac{63}{146}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

Concepts:random walk recursive probability symmetry

Difficulty rating: 2450

Small Hint:

By symmetry the escape probability from pad $5$ is $\tfrac12$

Big Hint:

Let $p_j$ be the escape probability from pad $j$ and write each $p_j$ in terms of its neighbors

Solution:

Let $p_j$ be the probability of eventually reaching pad $10$ starting from pad $j.$ By the symmetry of the jump rule at the center, $p_5 = \tfrac12.$

Each interior pad satisfies $p_j = \tfrac{10-j}{10}\,p_{j+1} + \tfrac{j}{10}\,p_{j-1},$ which gives $\begin{gathered} p_4 = \tfrac25 p_3 + \tfrac35 p_5, \\ \quad p_3 = \tfrac{3}{10} p_2 + \tfrac{7}{10} p_4, \end{gathered}$ $p_2 = \tfrac15 p_1 + \tfrac45 p_3,\quad p_1 = \tfrac{9}{10} p_2.$

Put $d_j=p_j-p_{j-1}.$ The recurrence is equivalent to $(10-j)d_{j+1}=jd_j,$ so $\begin{gathered} d_2=\dfrac{d_1}{9},\quad d_3=\dfrac{d_1}{36}, \\ d_4=\dfrac{d_1}{84},\quad d_5=\dfrac{d_1}{126}. \end{gathered}$ Since $p_0=0$ and $p_5=\frac{1}{2},$ $\begin{aligned} \dfrac12&=d_1\biggl(1+\dfrac19+\dfrac1{36} \\ &\qquad+\dfrac1{84}+\dfrac1{126}\biggr) \\ &=d_1\cdot\dfrac{73}{63}. \end{aligned}$ Therefore $p_1=d_1=\frac{63}{146}.$

Thus, the correct answer is C.

23.

The number $2017$ is prime. Let $S = \sum_{k=0}^{62} \binom{2014}{k}.$ What is the remainder when $S$ is divided by $2017?$

$32$

$684$

$1024$

$1576$

$2016$

Concepts:combinations modular arithmetic telescoping

Difficulty rating: 2560

Small Hint:

Modulo $2017,$ reduce $\binom{2014}{k}$ using $2016 \equiv -1,$ $2015 \equiv -2,$ and so on

Big Hint:

This gives $\binom{2014}{k} \equiv (-1)^k \binom{k+2}{2},$ and the alternating sum telescopes

Solution:

Working modulo $2017,$ the identity $\binom{2014}{k} \cdot k! \cdot (2014-k)! = 2014!$ together with $2016 \cdot 2015 \cdots (2015-k)$ $\equiv (-1)^k (k+2)!$ leads to $\begin{gathered} 2\binom{2014}{k} \equiv (-1)^k \\ {}\cdot (k+2)(k+1) \pmod{2017}, \end{gathered}$ so $\binom{2014}{k} \equiv (-1)^k \binom{k+2}{2}.$

Then $\begin{gathered} S \equiv \sum_{k=0}^{62} (-1)^k \binom{k+2}{2} \\ = 1 \\ {}+ \sum_{k=1}^{31}\left[\binom{2k+2}{2} - \binom{2k+1}{2}\right] \\ = 1 + \sum_{k=1}^{31}(2k+1). \end{gathered}$

The remaining sum is $3 + 5 + \cdots + 63 = 1023,$ so $S \equiv 1 + 1023$ $= 1024 \pmod{2017}.$

Thus, the correct answer is C.

24.

Let $ABCDE$ be a pentagon inscribed in a circle such that $AB = CD = 3,$ $BC = DE = 10,$ and $AE = 14.$ The sum of the lengths of all diagonals of $ABCDE$ is equal to $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m + n?$

$129$

$247$

$353$

$391$

$421$

Concepts:Ptolemy’s Theorem chord polynomial

Difficulty rating: 2650

Small Hint:

Equal chords subtend equal arcs, so $AC = BD = CE;$ call this $x$

Big Hint:

Apply Ptolemy’s theorem to $ABCD,$ $BCDE,$ and $ABDE$ to get equations in $x, y = AD, z = BE$

Solution:

Because arcs $AB, CD$ are equal and arcs $BC, DE$ are equal, the chords $AC, BD, CE$ are all equal; let $x = AC = BD = CE,$ $y = AD,$ and $z = BE.$

Ptolemy’s theorem on $ABCD,$ $BCDE,$ and $ABDE$ gives $\begin{gathered} 10y + 9 = x^2, \\ \quad 100 + 3z = x^2, \\ \quad 30 + 14x = yz. \end{gathered}$ Solving the first two for $y$ and $z$ and substituting into the third yields $\begin{gathered} x^3 - 109x - 420 = 0 \\ = (x-12)(x+5)(x+7). \end{gathered}$

So $x = 12,$ $y = \tfrac{135}{10} = \tfrac{27}{2},$ and $z = \tfrac{44}{3}.$ The five diagonals are $AC, BD, CE, AD, BE,$ summing to $\begin{gathered} 3x + y + z = 36 \\ {}+ \tfrac{27}{2} + \tfrac{44}{3} \\ = \tfrac{385}{6}. \end{gathered}$

Thus $m + n = 385 + 6 = 391,$ and the correct answer is D.

25.

What is the sum of all positive real solutions $x$ to the equation $\begin{gathered} \small 2\cos(2x)\left(\cos(2x) - \cos\left(\dfrac{2014\pi^2}{x}\right)\right) \\ = \cos(4x) - 1? \end{gathered}$

$\pi$

$810\pi$

$1008\pi$

$1080\pi$

$1800\pi$

Concepts:trigonometric identity sum of factors

Difficulty rating: 2890

Small Hint:

Substitute $x = \tfrac{\pi y}{2}$ and use $\tfrac12(1 - \cos(2\pi y)) = \sin^2(\pi y)$

Big Hint:

The equation reduces to $\cos(\pi y)\cos\left(\tfrac{4028\pi}{y}\right) = 1,$ forcing $y$ and $\tfrac{4028}{y}$ to be integers of the same parity

Solution:

Let $x = \tfrac{\pi y}{2}.$ Dividing by $2$ and using $\tfrac12(1 - \cos(2\pi y)) = \sin^2(\pi y),$ the equation simplifies to $\cos(\pi y)\cos\left(\dfrac{4028\pi}{y}\right) = 1.$

Both cosines must equal $1$ or both equal $-1,$ so $y$ and $\tfrac{4028}{y}$ are integers of the same parity. Since $4028 = 2^2 \cdot 19 \cdot 53$ is even, both must be even, so $y = 2a$ with $a$ a positive odd divisor of $2014 = 2 \cdot 19 \cdot 53,$ giving $a \in \{1, 19, 53, 19 \cdot 53\}.$

Each such $a$ gives $x = \tfrac{\pi y}{2} = \pi a,$ so the sum of solutions is $\begin{gathered} \pi(1 + 19 + 53 + 19\cdot53) \\ = \pi(19+1)(53+1) \\ = 1080\pi. \end{gathered}$

Thus, the correct answer is D.