2018 AMC 10B Solutions | LIVE by Po-Shen Loh

1.

Kate bakes a $20$ -inch by $18$ -inch pan of cornbread. The cornbread is cut into pieces that measure $2$ inches by $2$ inches. How many pieces of cornbread does the pan contain?

$90$

$100$

$180$

$200$

$360$

Concepts:area rectangle

Difficulty rating: 860

Small Hint:

The pan and each piece are rectangles; compare their areas

Big Hint:

Divide the total area by the area of one $2 \times 2$ piece

Solution:

The whole pan has area $20 \times 18 = 360$ square inches. Each piece is $2 \times 2 = 4$ square inches. So the number of pieces is $\frac{360}{4} = 90.$ Thus, A is the correct answer.

2.

Sam drove $96$ miles in $90$ minutes. His average speed during the first $30$ minutes was $60$ mph (miles per hour), and his average speed during the second $30$ minutes was $65$ mph. What was his average speed, in mph, during the last $30$ minutes?

$64$

$65$

$66$

$67$

$68$

Concepts:distance rate and time

Difficulty rating: 980

Small Hint:

Distance = rate $\times$ time, and each $30$ -minute leg is half an hour

Big Hint:

Subtract the distance covered in the first hour from $96$ to get the last leg’s distance

Solution:

Each leg is half an hour. In the first, Sam drove $60 \cdot \tfrac12 = 30$ miles; in the second, $65 \cdot \tfrac12 = 32.5$ miles. That’s $62.5$ miles so far. That leaves $96 - 62.5 = 33.5$ miles for the last half hour, which is a speed of $\frac{33.5}{\tfrac12} = 67$ mph. Therefore, the answer is D.

3.

In the expression $(\underline{\phantom{x}} \times \underline{\phantom{x}}) + (\underline{\phantom{x}} \times \underline{\phantom{x}})$ each blank is to be filled in with one of the digits $1,$ $2,$ $3,$ or $4,$ with each digit being used once. How many different values can be obtained?

$2$

$3$

$4$

$6$

$24$

Concepts:basic counting systematic listing

Difficulty rating: 950

Small Hint:

Neither multiplication nor the final addition depends on order, so only the way the four digits are split into two pairs matters

Big Hint:

List the three ways to split $\{1,2,3,4\}$ into two pairs

Solution:

Order inside a product doesn’t matter, and neither does the order we add the two products. So all that matters is how the four digits split into two pairs. There are three splits: $1\cdot2 + 3\cdot4 = 14,$ $1\cdot3 + 2\cdot4 = 11,$ and $1\cdot4 + 2\cdot3 = 10.$ That’s $3$ different values. Thus, B is the correct answer.

4.

A three-dimensional rectangular box with dimensions $X,$ $Y,$ and $Z$ has faces whose surface areas are $24,$ $24,$ $48,$ $48,$ $72,$ and $72$ square units. What is $X + Y + Z?$

$18$

$22$

$24$

$30$

$36$

Concepts:rectangular prism surface area system of equations

Difficulty rating: 1130

Small Hint:

The three distinct face areas are the pairwise products $XY, XZ, YZ,$ equal to $24, 48, 72$

Big Hint:

Multiplying all three gives $(XYZ)^2;$ take the square root, then divide by each face area

Solution:

The three distinct face areas are the pairwise products $XY = 24,$ $XZ = 48,$ $YZ = 72$ in some order. Multiply all three: $(XYZ)^2 = 24 \cdot 48 \cdot 72 = 82944,$ so $XYZ = 288.$ Now divide by each face area. We get $Z = \frac{288}{24} = 12,$ $Y = \frac{288}{48} = 6,$ and $X = \frac{288}{72} = 4,$ so $X + Y + Z = 22.$ Therefore, the answer is B.

5.

How many subsets of $\{2, 3, 4, 5, 6, 7, 8, 9\}$ contain at least one prime number?

$128$

$192$

$224$

$240$

$256$

Concepts:subsets complementary counting prime

Difficulty rating: 1200

Small Hint:

Count the subsets with no prime and subtract from the total number of subsets

Big Hint:

The non-prime elements are $4, 6, 8, 9;$ any subset of just those contains no prime

Solution:

Count the complement. The set has $2^8 = 256$ subsets total. A subset avoids every prime exactly when it sticks to the non-primes $\{4, 6, 8, 9\},$ and there are $2^4 = 16$ of those. So $256 - 16 = 240$ subsets contain at least one prime. Thus, D is the correct answer.

6.

A box contains $5$ chips, numbered $1,$ $2,$ $3,$ $4,$ and $5.$ Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds $4.$ What is the probability that $3$ draws are required?

$\dfrac{1}{15}$

$\dfrac{1}{10}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

Concepts:sampling without replacement basic probability

Difficulty rating: 1290

Small Hint:

A third draw is needed exactly when the first two chips still sum to at most $4$

Big Hint:

Which pairs of distinct chips sum to $4$ or less? Count them among all ordered first-two-draw outcomes

Solution:

We need a third draw exactly when the first two chips still sum to $4$ or less. The only such unordered pairs are $\{1,2\}$ and $\{1,3\}.$ Each can be drawn in either order, giving $4$ favorable ordered prefixes.

Imagine that a complete random ordering of all five chips is chosen in advance. Then all $5\cdot4=20$ ordered first-two-chip prefixes are equally likely, even when the actual process would stop after the first chip. Thus the probability is $\frac{4}{20}=\frac{1}{5}.$ Therefore, the answer is D.

7.

In the figure below, $N$ congruent semicircles are drawn along a diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let $A$ be the combined area of the small semicircles and $B$ be the area of the region inside the large semicircle but outside the small semicircles. The ratio $A : B$ is $1 : 18.$ What is $N?$

$16$

$17$

$18$

$19$

$36$

Concepts:circle area ratio and proportion

Difficulty rating: 1310

Small Hint:

If each small semicircle has radius $r,$ the large semicircle has radius $Nr$

Big Hint:

Show $A : B = 1 : (N-1),$ then set this equal to $1 : 18$

Solution:

Let each small semicircle have radius $r.$ The $N$ diameters cover the big diameter, so the large radius is $Nr.$ Then $A = N \cdot \tfrac12 \pi r^2,$ and the large semicircle has area $\tfrac12 \pi (Nr)^2,$ so the leftover region is $B = \tfrac12 \pi r^2(N^2 - N).$ This gives $A : B = N : N(N-1)$ $= 1 : (N-1).$ Set $N - 1 = 18,$ and $N = 19.$ Thus, D is the correct answer.

8.

Sara makes a staircase out of toothpicks as shown:

This is a $3$ -step staircase and uses $18$ toothpicks. How many steps would be in a staircase that used $180$ toothpicks?

$10$

$11$

$12$

$24$

$30$

Concepts:quadratic summation pattern recognition

Difficulty rating: 1200

Small Hint:

Count the toothpicks in an $n$ -step staircase; the vertical ones number $(1 + 2 + \cdots + n) + n$

Big Hint:

With equal horizontal and vertical counts the total is $n(n+3);$ solve $n(n+3) = 180$

Solution:

In an $n$ -step staircase the vertical toothpicks number $(1 + 2 + \cdots + n) + n$ $= \tfrac{n(n+1)}{2} + n,$ and there are just as many horizontal ones. That’s a total of $n(n+1) + 2n = n(n+3).$ Check: $n = 3$ gives $18,$ as it should. Now solve $n(n+3) = 180.$ This factors as $(n-12)(n+15) = 0,$ so $n = 12.$ Therefore, the answer is C.

9.

The faces of each of $7$ standard dice are labeled with the integers from $1$ to $6.$ Let $p$ be the probability that when all $7$ dice are rolled, the sum of the numbers on the top faces is $10.$ What other sum occurs with the same probability $p?$

$13$

$26$

$32$

$39$

$42$

Concepts:dice (probability)bijection symmetry

Difficulty rating: 1370

Small Hint:

Replacing each die value $k$ by $7 - k$ pairs up equally likely outcomes

Big Hint:

This replacement sends a total of $s$ to a total of $49 - s$

Solution:

Replace each die’s value $k$ by $7 - k.$ This pairs up outcomes one-to-one and keeps their probabilities, and it sends a total of $s$ to $7 \cdot 7 - s = 49 - s.$ So the sums $s$ and $49 - s$ are equally likely. The partner of $10$ is $49 - 10 = 39.$ Thus, D is the correct answer.

10.

In the rectangular parallelepiped shown, $AB = 3,$ $BC = 1,$ and $CG = 2.$ Point $M$ is the midpoint of $FG.$ What is the volume of the rectangular pyramid with base $BCHE$ and apex $M?$

$1$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$2$

Concepts:volume pyramid coordinate geometry

Difficulty rating: 1570

Small Hint:

Place $A$ at the origin with the edges along the axes; find the rectangle $BCHE$ and its area

Big Hint:

Volume $= \tfrac13$ $\times (\text{base area})$ $\times (\text{distance from } M \text{ to base})$

Solution:

Put $A$ at the origin with edges along the axes: $A = (0,0,0),$ $B = (3,0,0),$ $C = (3,1,0),$ $E = (0,0,2),$ $H = (0,1,2),$ $F = (3,0,2),$ $G = (3,1,2),$ so $M = (3, \tfrac12, 2).$ The base $BCHE$ is a rectangle with $BC = 1$ and $BE = \sqrt{3^2 + 2^2} = \sqrt{13},$ hence area $\sqrt{13}.$ Its plane is $2x + 3z = 6,$ and $M$ sits at distance $\frac{|2\cdot3 + 3\cdot2 - 6|}{\sqrt{13}} = \frac{6}{\sqrt{13}}$ from it. The volume is $\tfrac13 \cdot \sqrt{13} \cdot \frac{6}{\sqrt{13}} = 2.$ Therefore, the answer is E.

11.

Which of the following expressions is never a prime number when $p$ is a prime number?

$p^2 + 16$

$p^2 + 24$

$p^2 + 26$

$p^2 + 46$

$p^2 + 96$

Concepts:prime modular arithmetic

Difficulty rating: 1500

Small Hint:

Test the expressions modulo $3;$ for a prime $p \ne 3,$ $p^2 \equiv 1 \pmod 3$

Big Hint:

For the correct choice, also check $p = 3$ separately, then show it is a multiple of $3$ for all other primes

Solution:

Look at $p^2 + 26.$ When $p = 3,$ it’s $35 = 5 \cdot 7.$ For any other prime, $p$ isn’t divisible by $3,$ so $p^2 \equiv 1 \pmod 3$ and $p^2 + 26 \equiv 1 + 2 \equiv 0 \pmod 3.$ Either way it’s a multiple of $3$ bigger than $3,$ hence composite. So it’s never prime. Thus, C is the correct answer.

12.

Line segment $AB$ is a diameter of a circle with $AB = 24.$ Point $C,$ not equal to $A$ or $B,$ lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?

$25$

$38$

$50$

$63$

$75$

Concepts:centroid homothety circle area

Difficulty rating: 1530

Small Hint:

The centroid of $\triangle ABC$ is $\tfrac13(A + B + C),$ and $A + B$ is twice the center

Big Hint:

As $C$ moves on a circle of radius $12,$ the centroid moves on a circle of one third the radius

Solution:

Put the center $O$ at the origin, so $A = (-12, 0)$ and $B = (12, 0),$ while $C$ runs over the circle of radius $12.$ Then $A + B = 0,$ so the centroid is $\tfrac13(A + B + C) = \tfrac13 C.$ As $C$ circles, $\tfrac13 C$ traces a circle of radius $\tfrac{12}{3} = 4$ (minus the two points where $C = A$ or $B$ ). Its area is $\pi \cdot 4^2 = 16\pi \approx 50.$ Therefore, the answer is C.

13.

How many of the first $2018$ numbers in the sequence $101,$ $1001,$ $10001,$ $100001,$ $\ldots$ are divisible by $101?$

$253$

$504$

$505$

$506$

$1009$

Concepts:modular arithmetic multiplicative order counting integers in a range

Difficulty rating: 1570

Small Hint:

The $k$ -th term is $10^{k+1} + 1,$ which is divisible by $101$ when $10^{k+1} \equiv -1 \pmod{101}$

Big Hint:

$10^2 = 100 \equiv -1 \pmod{101},$ so $10^m \equiv -1$ exactly when $m \equiv 2 \pmod 4$

Solution:

The $k$ -th term is $10^{k+1} + 1,$ which $101$ divides iff $10^{k+1} \equiv -1 \pmod{101}.$ Notice $10^2 = 100 \equiv -1 \pmod{101}.$ So $10^m \equiv -1$ exactly when $m \equiv 2 \pmod 4,$ meaning $k + 1 \equiv 2,$ that is $k \equiv 1 \pmod 4.$ Among $k = 1, 2, \ldots, 2018,$ the values $1, 5, \ldots, 2017$ number $505.$ Thus, C is the correct answer.

14.

A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?

$202$

$223$

$224$

$225$

$234$

Concepts:mode optimization

Difficulty rating: 1660

Small Hint:

Every value other than the mode can appear at most $9$ times, to keep the mode unique

Big Hint:

With $d$ distinct values the list holds at most $10 + 9(d-1)$ entries; make this at least $2018$

Solution:

The mode shows up $10$ times. To keep the number of distinct values small, let every other value repeat as much as the rules allow, which is $9$ times each (any more would tie the mode). With $d$ distinct values the list holds at most $10 + 9(d-1)$ entries. We need $10 + 9(d-1) \ge 2018,$ so $d - 1 \ge 223.1,$ giving $d \ge 225.$ Therefore, the answer is D.

15.

A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h.$ What is the area of the sheet of wrapping paper?

$2(w + h)^2$

$\dfrac{(w + h)^2}{2}$

$2w^2 + 4wh$

$2w^2$

$w^2 h$

Concepts:paper folding Pythagorean Theorem square (geometry)

Difficulty rating: 1730

Small Hint:

If the sheet has side $s,$ its center-to-corner distance is $\dfrac{s}{\sqrt 2}$

Big Hint:

Folding a corner to the top center covers, in a straight line, half the base, then the side of height $h,$ then the other half of the top

Solution:

Let the sheet have side $s.$ The base sits as a square of side $w$ turned $45^\circ,$ so the center is $\tfrac{w}{2}$ from each base edge. A corner of the sheet lies $\tfrac{s}{\sqrt2}$ from the center. Folding that corner up to the top center traces a straight line: $\tfrac{w}{2}$ out to the base edge, then $h$ up the side, then $\tfrac{w}{2}$ across the top. So $\tfrac{s}{\sqrt2} = \tfrac{w}{2} + h + \tfrac{w}{2} = w + h.$ Then $s = \sqrt2\,(w + h),$ and the area is $s^2 = 2(w + h)^2.$ Thus, A is the correct answer.

16.

Let $a_1,$ $a_2,$ $\ldots,$ $a_{2018}$ be a strictly increasing sequence of positive integers such that

$a_1 + a_2 + \cdots + a_{2018} = 2018^{2018}.$

What is the remainder when $a_1^3 + a_2^3 + \cdots + a_{2018}^3$ is divided by $6?$

$0$

$1$

$2$

$3$

$4$

Concepts:modular arithmetic divisibility

Difficulty rating: 1710

Small Hint:

$n^3 \equiv n \pmod 6,$ because $n^3 - n = (n-1)n(n+1)$ is a product of three consecutive integers

Big Hint:

So the sum of cubes is congruent to $2018^{2018} \pmod 6;$ reduce that modulo $6$

Solution:

For any integer $n,$ $n^3 - n = (n-1)n(n+1)$ is a product of three consecutive integers, so it’s divisible by $6.$ That means $n^3 \equiv n \pmod 6.$ Summing, $\sum a_i^3 \equiv \sum a_i$ $= 2018^{2018} \pmod 6.$ Now $2018 \equiv 2 \pmod 6,$ and powers of $2$ mod $6$ alternate $2, 4, 2, 4, \ldots.$ The exponent $2018$ is even, so $2^{2018} \equiv 4 \pmod 6.$ The remainder is $4.$ Therefore, the answer is E.

17.

In rectangle $PQRS,$ $PQ = 8$ and $QR = 6.$ Points $A$ and $B$ lie on $PQ,$ points $C$ and $D$ lie on $QR,$ points $E$ and $F$ lie on $RS,$ and points $G$ and $H$ lie on $SP$ so that $AP = BQ < 4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k + m\sqrt{n},$ where $k,$ $m,$ and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k + m + n?$

$1$

$7$

$21$

$92$

$106$

Concepts:Pythagorean Theorem quadratic

Difficulty rating: 1890

Small Hint:

The four cut corners are congruent right triangles with legs $x$ (on the sides of length $8$ ) and $y$ (on the sides of length $6$ )

Big Hint:

The equal sides give $8 - 2x = 6 - 2y = \sqrt{x^2 + y^2};$ the first equality gives $y = x - 1,$ then substitute and square

Solution:

Let $s$ be the octagon’s side length and let $x=AP=BQ=\frac{8-s}{2}.$ The right triangles $\triangle APH$ and $\triangle BQC$ have the same hypotenuse $s$ and a leg of length $x,$ so they are congruent; write $PH=QC=y.$ Because $CD=HG=s$ and the vertical sides of the rectangle both have length $6,$ it follows that $DR=GS.$ The right triangles at $R$ and $S$ are then congruent, so $RE=SF.$ Since $RS=8$ and $EF=s,$ each of these equal lengths is $\frac{8-s}{2}=x.$ Thus all four cut corners have legs $x$ and $y.$

The equal octagon sides give $8-2x=6-2y=\sqrt{x^2+y^2}.$ The first equality gives $y=x-1.$ Substituting and squaring gives $2x^2-30x+63=0,$ so the root with $x<4$ is $x=\frac{15-3\sqrt{11}}{2}.$

The side length is $8-2x=-7+3\sqrt{11},$ so $k+m+n=-7+3+11=7.$ Thus, B is the correct answer.

18.

Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?

$60$

$72$

$92$

$96$

$120$

Concepts:casework permutations derangements

Difficulty rating: 1930

Small Hint:

Show that each row must contain exactly one child from each family

Big Hint:

The third row’s families must differ from the second row’s in every column (a derangement), and each pair’s two children can be swapped between their seats

Solution:

Suppose some family put both children in one row. They’d have to take the non-adjacent seats $1$ and $3,$ which forces the middle family’s two children into the same column. Not allowed. So each row holds exactly one child from each family. The second row is a permutation of the three families, $3! = 6$ ways. The third row needs a different family in every column, a derangement of the second row’s order, and there are $2$ of those. Finally, each pair can swap its two children between their seats, $2^3 = 8$ ways. The total is $6 \cdot 2 \cdot 8 = 96.$ Therefore, the answer is D.

19.

Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe’s age will be an integral multiple of Zoe’s age. What will be the sum of the two digits of Joey’s age the next time his age is a multiple of Zoe’s age?

$7$

$8$

$9$

$10$

$11$

Concepts:factor counting divisibility digits

Difficulty rating: 1990

Small Hint:

Let Chloe be $n$ today. In $t$ years her age is a multiple of Zoe’s iff $1 + t$ divides $n - 1$

Big Hint:

Exactly $9$ such birthdays means $n - 1$ has $9$ divisors; find the two-digit value, then repeat the divisor idea for Joey

Solution:

Let Chloe be $n$ today; Zoe is $1.$ In $t$ years their age ratio is $\dfrac{n+t}{1+t}=1+\dfrac{n-1}{1+t},$ which is an integer exactly when $1+t$ divides $n-1.$ Thus $n-1$ has exactly $9$ positive divisors.

A number with $9$ divisors has the form $p^8$ or $p^2q^2.$ The only two-digit possibility is $2^2\cdot3^2=36,$ so Chloe is $37$ and Joey is $38.$

Joey’s age $38+t$ is a multiple of Zoe’s age $1+t$ exactly when $1+t$ divides $37.$ The next time is $t=36,$ when Joey is $74.$ Its digit sum is $7+4=11.$ Thus, E is the correct answer.

20.

A function $f$ is defined recursively by $f(1) = f(2) = 1$ and

$f(n) = f(n - 1) - f(n - 2) + n$

for all integers $n \ge 3.$ What is $f(2018)?$

$2016$

$2017$

$2018$

$2019$

$2020$

Concepts:recursion pattern recognition

Difficulty rating: 1910

Small Hint:

Look for a particular solution of the form $f(n) = n + c;$ subtracting it leaves $g(n) = g(n-1) - g(n-2)$

Big Hint:

That homogeneous recursion is periodic with period $6;$ reduce $2018$ modulo $6$

Solution:

Notice $f(n) = n + 1$ solves the recurrence on its own, so write $f(n) = (n + 1) + g(n).$ Then $g$ satisfies the homogeneous version $g(n) = g(n-1) - g(n-2).$ With $g(1) = -1$ and $g(2) = -2,$ it cycles with period $6$ : $-1, -2, -1, 1, 2, 1, \ldots.$ Since $2018 \equiv 2 \pmod 6,$ we get $g(2018) = -2,$ so $f(2018) = 2019 - 2 = 2017.$ Therefore, the answer is B.

21.

Mary chose an even $4$ -digit number $n.$ She wrote down all the divisors of $n$ in increasing order from left to right: $1,$ $2,$ $\ldots,$ $\frac{n}{2},$ $n.$ At some moment Mary wrote $323$ as a divisor of $n.$ What is the smallest possible value of the next divisor written to the right of $323?$

$324$

$330$

$340$

$361$

$646$

Concepts:factor least common multiple prime factorization

Difficulty rating: 2100

Small Hint:

$323 = 17 \cdot 19,$ and $n$ is an even $4$ -digit multiple of $323$

Big Hint:

For a candidate next divisor $d,$ $n$ must be a multiple of $\operatorname{lcm}(323, d);$ find the smallest $d > 323$ keeping that under $10000$

Solution:

Let $d$ be the next divisor after $323.$ If $\gcd(d,323)=1,$ then $n$ is a multiple of $323d>323^2>9999,$ impossible for a four-digit number. Thus $\gcd(d,323)>1.$

Since $323=17\cdot19,$ this gcd is at least $17.$ It also divides $d-323,$ so $d-323\ge17$ and hence $d\ge340.$

This bound is attained: for $n=6460=2^2\cdot5\cdot17\cdot19,$ both $323$ and $340$ are divisors. The lower bound shows there is no divisor between them. Thus the smallest possible next divisor is $340,$ and C is the correct answer.

22.

Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $[0, 1].$ Which of the following numbers is closest to the probability that $x,$ $y,$ and $1$ are the side lengths of an obtuse triangle?

$0.21$

$0.25$

$0.29$

$0.50$

$0.79$

Concepts:geometric probability law of cosines

Difficulty rating: 2100

Small Hint:

A triangle with sides $x, y, 1$ needs $x + y > 1;$ since $1$ is the longest side it is obtuse when $x^2 + y^2 < 1$

Big Hint:

In the unit square this region is a quarter disk with the triangle below $x + y = 1$ removed

Solution:

The three lengths $x, y, 1$ make a triangle iff $x + y > 1.$ Since $1$ is the longest side, that triangle is obtuse iff $x^2 + y^2 < 1.$ So in the unit square we want the region inside the quarter circle $x^2 + y^2 = 1$ but above the line $x + y = 1.$ That’s the quarter disk with the right triangle under the chord removed: $\tfrac{\pi}{4} - \tfrac12 \approx 0.285.$ The closest choice is $0.29.$ Therefore, the answer is C.

23.

How many ordered pairs $(a, b)$ of positive integers satisfy the equation

$\begin{aligned} a \cdot b + 63 &= 20 \cdot \operatorname{lcm}(a, b) \\ &\quad {}+ 12 \cdot \gcd(a, b), \end{aligned}$

where $\gcd(a, b)$ denotes the greatest common divisor of $a$ and $b,$ and $\operatorname{lcm}(a, b)$ denotes their least common multiple?

$0$

$2$

$4$

$6$

$8$

Concepts:least common multiple greatest common divisor Simon’s Favorite Factoring Trick

Difficulty rating: 2120

Small Hint:

Use $ab = \gcd(a,b) \cdot \operatorname{lcm}(a,b);$ let $x = \operatorname{lcm}(a,b)$ and $y = \gcd(a,b)$

Big Hint:

The equation becomes $(x - 12)(y - 20) = 177 = 3 \cdot 59;$ also $y$ must divide $x$

Solution:

Recall $ab=\gcd(a,b)\operatorname{lcm}(a,b).$ Let $x=\operatorname{lcm}(a,b)$ and $y=\gcd(a,b).$ The equation becomes $(x-12)(y-20)=177=3\cdot59.$

The positive factor pairs give $(x,y)=(13,197),$ $(189,21),$ $(15,79),$ and $(71,23).$ The negative factor pairs make either $x$ or $y$ negative, so they are impossible. Also $y$ must divide $x,$ and only $(x,y)=(189,21)$ passes.

Write $a=21u$ and $b=21v.$ Then $\gcd(u,v)=1$ and $uv=\frac{189}{21}=9,$ so $(u,v)=(1,9)$ or $(9,1).$ Hence the two ordered pairs are $(21,189)$ and $(189,21),$ and B is the correct answer.

24.

Let $ABCDEF$ be a regular hexagon with side length $1.$ Denote by $X,$ $Y,$ and $Z$ the midpoints of sides $AB,$ $CD,$ and $EF,$ respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ?$

$\dfrac{3}{8}\sqrt{3}$

$\dfrac{7}{16}\sqrt{3}$

$\dfrac{15}{32}\sqrt{3}$

$\dfrac{1}{2}\sqrt{3}$

$\dfrac{9}{16}\sqrt{3}$

Concepts:regular polygon equilateral triangle area decomposition

Difficulty rating: 2470

Small Hint:

$\triangle ACE$ is equilateral with side $\sqrt{3};$ $\triangle XYZ$ is equilateral with side $\tfrac{3}{2}$

Big Hint:

Centered at the hexagon’s center the two triangles are concentric and rotated $30^\circ;$ the overlap is $\triangle XYZ$ minus its three protruding corner triangles

Solution:

The triangle $XYZ$ is equilateral with side $\frac{3}{2},$ so its area is $\dfrac{\sqrt3}{4}\left(\dfrac32\right)^2=\dfrac{9\sqrt3}{16}.$

The triangles $ACE$ and $XYZ$ are concentric and rotated $30^\circ$ from each other. At each vertex of $XYZ,$ the sides of $ACE$ cut off a $30$ - $60$ - $90$ triangle whose hypotenuse is the half-side segment $AX=\frac{1}{2}.$ Its legs are $\frac{1}{4}$ and $\frac{\sqrt3}{4},$ so each corner has area $\frac{\sqrt3}{32}.$

Removing the three corners gives $\dfrac{9\sqrt3}{16}-3\cdot\dfrac{\sqrt3}{32}=\dfrac{15\sqrt3}{32}.$ Therefore, the answer is C.

25.

Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ How many real numbers $x$ satisfy the equation $x^2 + 10{,}000\lfloor x \rfloor = 10{,}000x?$

$197$

$198$

$199$

$200$

$201$

Concepts:floor and ceiling functions counting integers in a range

Difficulty rating: 2270

Small Hint:

Writing $\{x\} = x - \lfloor x \rfloor,$ the equation is $x^2 = 10{,}000\{x\},$ so $0 \le x^2 < 10{,}000$

Big Hint:

On each interval $[a, a+1)$ there is a solution exactly when $(a + 1)^2 \lt 10{,}000$

Solution:

Let $a = \lfloor x \rfloor.$ The equation reads $x^2 = 10{,}000(x - a)$ $= 10{,}000\{x\},$ and since $0 \le \{x\} < 1,$ this forces $0 \le x^2 < 10{,}000,$ so $-100 < x < 100.$ On each interval $[a, a + 1)$ the quantity $10{,}000x - x^2$ increases from $10{,}000a - a^2$ and approaches, but does not reach, $10{,}000(a+1) - (a+1)^2.$ It hits $10{,}000a$ exactly once precisely when $(a + 1)^2 < 10{,}000.$ That holds for the integers $-100 \le a \le 98,$ which is $199$ solutions. Thus, C is the correct answer.