2020 AMC 12A Solutions | LIVE by Po-Shen Loh

1.

Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?

$10\%$

$15\%$

$20\%$

$30\%$

$35\%$

Concepts:percentage fraction

Difficulty rating: 840

Small Hint:

After Carlos takes his share, $30\%$ of the pie remains

Big Hint:

Maria removes $\tfrac13$ of that $30\%$ , so $\tfrac23$ of the remainder is left

Solution:

After Carlos takes $70\%,$ the remaining portion is $30\%.$

Maria takes one third of this, namely $\tfrac13 \cdot 30\% = 10\%,$ leaving $30\% - 10\% = 20\%.$

Thus, C is the correct answer.

2.

The acronym AMC is shown in the rectangular grid below with grid lines spaced $1$ unit apart. In units, what is the sum of the lengths of the line segments that form the acronym AMC?

$17$

$15 + 2\sqrt{2}$

$13 + 4\sqrt{2}$

$11 + 6\sqrt{2}$

$21$

Concepts:Pythagorean Theorem distance formula

Difficulty rating: 1020

Small Hint:

A segment spanning one unit right and one unit up has length $\sqrt{2}$

Big Hint:

Total the horizontal and vertical pieces separately from the diagonal pieces

Solution:

Split each letter into its segments. The $A$ is a diagonal of length $2\sqrt{2},$ a vertical of length $2,$ and a crossbar of length $1.$

The $M$ has two verticals of length $2$ and two diagonals of length $\sqrt{2}$ each. The $C$ is three sides of length $2.$

The straight pieces total $2 + 1 + 2 + 2 + 2 + 2 + 2 = 13,$ and the diagonal pieces total $2\sqrt2 + \sqrt2 + \sqrt2 = 4\sqrt2.$

The sum is $13 + 4\sqrt2.$

Thus, C is the correct answer.

3.

A driver travels for $2$ hours at $60$ miles per hour, during which her car gets $30$ miles per gallon of gasoline. She is paid $\$0.50$ per mile, and her only expense is gasoline at $\$2.00$ per gallon. What is her net rate of pay, in dollars per hour, after this expense?

$20$

$22$

$24$

$25$

$26$

Concepts:rate unit conversion

Difficulty rating: 1130

Small Hint:

In $2$ hours she drives $120$ miles; find her pay and her gasoline cost for that distance

Big Hint:

Divide the net earnings by the $2$ hours to get the rate per hour

Solution:

In $2$ hours she drives $120$ miles, earning $120 \cdot \$0.50 = \$60.$

She uses $120 \div 30 = 4$ gallons, costing $4 \cdot \$2.00 = \$8.$

Her net earnings are $\$60 - \$8 = \$52,$ so her rate is $\$52 \div 2 = \$26$ per hour.

Thus, E is the correct answer.

4.

How many $4$ -digit positive integers (that is, integers between $1000$ and $9999,$ inclusive) having only even digits are divisible by $5?$

$80$

$100$

$125$

$200$

$500$

Concepts:multiplication principle divisibility digits

Difficulty rating: 1200

Small Hint:

A number divisible by $5$ ends in $0$ or $5,$ but the last digit must also be even

Big Hint:

Count the choices for each digit: the leading digit cannot be $0$

Solution:

To be divisible by $5$ the last digit is $0$ or $5,$ and to be even it must be $0.$ So the units digit is fixed.

The leading digit is a nonzero even digit: $2, 4, 6, 8$ give $4$ choices. Each of the two middle digits is any even digit $0, 2, 4, 6, 8,$ giving $5$ choices each.

The total is $4 \cdot 5 \cdot 5 \cdot 1 = 100.$

Thus, B is the correct answer.

5.

The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?

$2$

$5$

$10$

$25$

$50$

Concepts:magic square arithmetic sequence

Difficulty rating: 1130

Small Hint:

The total of all $25$ numbers is split equally among the $5$ rows

Big Hint:

Sum the arithmetic series from $-10$ to $14,$ then divide by $5$

Solution:

The sum of the $25$ integers is $\dfrac{(-10 + 14) \cdot 25}{2} = 50.$

The five rows each have the same sum and together account for the total, so each row sums to $50 \div 5 = 10.$

Thus, C is the correct answer.

6.

In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?

$4$

$5$

$6$

$7$

$8$

Concepts:symmetry

Difficulty rating: 1270

Small Hint:

The two symmetry lines must be the central vertical and central horizontal lines of the rectangle

Big Hint:

Each shaded square forces its reflections across both lines to be shaded too

Solution:

For both symmetries, the lines must be the vertical and horizontal center lines of the $5$ -by- $4$ grid. Every shaded square then forces the squares obtained by reflecting it across each line.

The top square lies off-center, so its reflection group has $4$ squares, requiring $3$ more. The middle square sits on the central column, so its group has $2$ squares, requiring $1$ more. The bottom-right square again has a group of $4,$ requiring $3$ more.

The least number of additional squares is $3 + 1 + 3 = 7.$

Thus, D is the correct answer.

7.

Seven cubes, whose volumes are $1,$ $8,$ $27,$ $64,$ $125,$ $216,$ and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?

$644$

$658$

$664$

$720$

$749$

Concepts:surface area sum of first n squares

Difficulty rating: 1340

Small Hint:

The cubes have side lengths $1, 2, \ldots, 7;$ handle the four vertical faces and the horizontal faces separately

Big Hint:

Looking straight down, all upward-facing horizontal area equals the base’s $7 \times 7;$ the same holds looking up

Solution:

The side lengths are $1, 2, \ldots, 7.$ The four side faces of cube $k$ contribute $4k^2,$ so the vertical faces total $4(1^2 + 2^2 + \cdots + 7^2)$ $= 4 \cdot 140 = 560.$

Viewed from directly above, every upward-facing horizontal patch projects onto the $7 \times 7$ base without overlap, giving $49.$ Viewed from below, the same is true, giving another $49.$

The total surface area is $560 + 49 + 49 = 658.$

Thus, B is the correct answer.

8.

What is the median of the following list of $4040$ numbers?

$1,$ $2,$ $3,$ $\ldots,$ $2020,$ $1^2,$ $2^2,$ $3^2,$ $\ldots,$ $2020^2$

$1974.5$

$1975.5$

$1976.5$

$1977.5$

$1978.5$

Concepts:median (data)perfect square counting integers in a range

Difficulty rating: 1440

Small Hint:

With $4040$ numbers the median averages the $2020$ th and $2021$ st smallest

Big Hint:

Squares up to $2020$ come from $1^2$ through $44^2;$ count how many list entries are at most a given integer

Solution:

The median is the average of the $2020$ th and $2021$ st smallest values.

The perfect squares that are at most $2020$ are $1^2, \ldots, 44^2$ (since $44^2 = 1936$ and $45^2 = 2025$ ), so there are $44$ of them.

Among the list, the numbers $\le 1976$ are the $1976$ integers $1, \ldots, 1976$ together with those $44$ squares, totaling $1976 + 44 = 2020.$

Thus the $2020$ th value is $1976$ and the $2021$ st value is $1977,$ making the median $\dfrac{1976 + 1977}{2} = 1976.5.$

Thus, C is the correct answer.

9.

How many solutions does the equation $\tan(2x) = \cos\left(\dfrac{x}{2}\right)$ have on the interval $[0, 2\pi]?$

$1$

$2$

$3$

$4$

$5$

Concepts:trigonometry

Difficulty rating: 1560

Small Hint:

On $[0, 2\pi],$ $\cos\left(\tfrac{x}{2}\right)$ decreases steadily from $1$ to $-1$

Big Hint:

$\tan(2x)$ has period $\tfrac{\pi}{2};$ count how many of its branches cross that decreasing curve

Solution:

On $[0, 2\pi],$ the graph of $\cos\left(\tfrac{x}{2}\right)$ is a single arc decreasing from $1$ down to $-1.$

The function $\tan(2x)$ has period $\tfrac{\pi}{2}$ with vertical asymptotes at $x = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}.$ These split the interval into five branches. On every branch $\tan(2x)$ is strictly increasing, while $\cos(\tfrac{x}{2})$ is decreasing, so there is at most one intersection per branch.

Each of the three interior branches runs from $-\infty$ to $+\infty,$ so each has one intersection. On the first branch, $\tan(0)=0\lt1=\cos(0)$ and the tangent tends to $+\infty.$ On the last, the tangent starts at $-\infty$ and ends at $0\gt-1=\cos\pi.$ Thus the two outer branches also have one intersection each, for $5$ total.

Thus, E is the correct answer.

10.

There is a unique positive integer $n$ such that $\log_2(\log_{16} n) = \log_4(\log_4 n).$

What is the sum of the digits of $n?$

$4$

$7$

$8$

$11$

$13$

Concepts:logarithm substitution

Difficulty rating: 1500

Small Hint:

$\log_{16} n = \tfrac12 \log_4 n,$ so let $y = \log_4 n$

Big Hint:

The equation becomes $\log_2\left(\tfrac{y}{2}\right) = \tfrac12 \log_2 y;$ square to solve for $y$

Solution:

Since $\log_{16} n = \tfrac12 \log_4 n,$ set $y = \log_4 n.$ The equation becomes $\log_2\left(\tfrac{y}{2}\right) = \log_4 y = \tfrac12 \log_2 y.$

Multiplying by $2$ gives $\log_2\left(\tfrac{y}{2}\right)^2 = \log_2 y,$ so $\left(\tfrac{y}{2}\right)^2 = y,$ which yields $y = 4.$

Then $\log_4 n = 4,$ so $n = 4^4 = 256,$ and the digit sum is $2 + 5 + 6 = 13.$

Thus, E is the correct answer.

11.

A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1,$ and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0, 0),$ $(0, 4),$ $(4, 4),$ and $(4, 0).$ What is the probability that the sequence of jumps ends on a vertical side of the square?

$\dfrac{1}{2}$

$\dfrac{5}{8}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{7}{8}$

Concepts:random walk system of equations symmetry

Difficulty rating: 1630

Small Hint:

Let $P(x, y)$ be the probability of ending on a vertical side; it equals the average of $P$ at the four neighbors

Big Hint:

Use symmetry: $P(2, 2) = \tfrac12,$ and the interior points reduce to a small system of equations

Solution:

Let $P(x, y)$ be the probability of ending on a vertical side. On a vertical side $P = 1,$ on a horizontal side $P = 0,$ and at an interior point $P$ is the average of its four neighbors.

By left-right symmetry $P(2, 2) = \tfrac12.$ Let $a = P(1, 2),$ $b = P(1, 1) = P(1, 3),$ and $c = P(2, 1) = P(2, 3).$ Then

$a = \tfrac14\left(1 + \tfrac12 + 2b\right),$ $\;b = \tfrac14(1 + c + a),$ and $c = \tfrac14\left(2b + \tfrac12\right).$

Substituting gives $b = \tfrac12,$ hence $a = \tfrac38 + \tfrac12 b = \tfrac58.$

Thus, B is the correct answer.

12.

Line $\ell$ in the coordinate plane has the equation $3x - 5y + 40 = 0.$ This line is rotated $45^\circ$ counterclockwise about the point $(20, 20)$ to obtain line $k.$ What is the $x$ -coordinate of the $x$ -intercept of line $k?$

$10$

$15$

$20$

$25$

$30$

Concepts:slope transformation coordinate geometry

Difficulty rating: 1630

Small Hint:

The point $(20, 20)$ lies on $\ell,$ so it also lies on $k;$ find the new slope

Big Hint:

Rotating a slope $m$ by $45^\circ$ gives $\dfrac{m + 1}{1 - m}$

Solution:

Note $(20, 20)$ satisfies $3x - 5y + 40 = 0,$ so it is on $\ell$ and remains on $k.$ The slope of $\ell$ is $\tfrac{3}{5}.$

Rotating by $45^\circ$ gives slope $\dfrac{\tfrac35 + 1}{1 - \tfrac35} = \dfrac{\tfrac85}{\tfrac25} = 4.$

Line $k$ is $y - 20 = 4(x - 20).$ Setting $y = 0$ gives $-20 = 4(x - 20),$ so $x - 20 = -5$ and $x = 15.$

Thus, B is the correct answer.

13.

There are integers $a,$ $b,$ and $c,$ each greater than $1,$ such that $\sqrt[a]{N \sqrt[b]{N \sqrt[c]{N}}} = \sqrt[36]{N^{25}}$ for all $N \gt 1.$ What is $b?$

$2$

$3$

$4$

$5$

$6$

Concepts:radical exponent

Difficulty rating: 1590

Small Hint:

Write the left side as $N$ to the power $\dfrac{1}{a} + \dfrac{1}{ab} + \dfrac{1}{abc}$

Big Hint:

Set that exponent equal to $\dfrac{25}{36}$ and peel off one nested variable at a time starting with $a = 2$

Solution:

The left side equals $N$ raised to the exponent $\dfrac{1}{a} + \dfrac{1}{ab} + \dfrac{1}{abc},$ which must equal $\dfrac{25}{36}.$

Because $b,c\ge2,$ this exponent is at most $\dfrac{7}{4a}.$ If $a\ge3,$ it is at most $\dfrac{7}{12}\lt\dfrac{25}{36},$ so $a=2.$ The equation becomes $\dfrac{1}{2b}+\dfrac{1}{2bc}=\dfrac{7}{36}.$

Clearing denominators gives $c(7b-18)=18.$ Set $d=7b-18,$ so $d$ is a positive divisor of $18.$ Among $1,2,3,6,9,18,$ only $d=3$ makes $b=\dfrac{d+18}{7}$ an integer. Hence $b=3$ and $c=6.$

Thus, B is the correct answer.

14.

Regular octagon $ABCDEFGH$ has area $n.$ Let $m$ be the area of quadrilateral $ACEG.$ What is $\dfrac{m}{n}?$

$\dfrac{\sqrt{2}}{4}$

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

$\dfrac{3}{4}$

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

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

Concepts:regular polygon area ratio

Difficulty rating: 1690

Small Hint:

$ACEG$ is a square inscribed in the octagon

Big Hint:

Place the octagon symmetrically and compute both areas; the four corner triangles make up the difference

Solution:

The four vertices $A, C, E, G$ form a square, since they are every other vertex of the regular octagon.

Taking a unit circumradius, the octagon’s area is $2\sqrt2$ and the square $ACEG$ has diagonal equal to the circle’s diameter, giving area $2.$

The ratio is $\dfrac{2}{2\sqrt2} = \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2}.$

Thus, B is the correct answer.

15.

In the complex plane, let $A$ be the set of solutions to $z^3 - 8 = 0$ and let $B$ be the set of solutions to $z^3 - 8z^2 - 8z + 64 = 0.$ What is the greatest distance between a point of $A$ and a point of $B?$

$2\sqrt{3}$

$6$

$9$

$2\sqrt{21}$

$9 + \sqrt{3}$

Concepts:complex number factoring distance formula

Difficulty rating: 1690

Small Hint:

$A$ is the three cube roots of $8;$ factor the second cubic by grouping

Big Hint:

The largest distance pairs a nonreal cube root of $8$ with the farthest real root of the second equation

Solution:

The set $A$ consists of the cube roots of $8:$ $2,$ $-1 + i\sqrt3,$ and $-1 - i\sqrt3.$

Factoring by grouping, $z^3 - 8z^2 - 8z + 64$ $= z^2(z - 8) - 8(z - 8)$ $= (z - 8)(z^2 - 8),$ so $B = \{8, 2\sqrt2, -2\sqrt2\},$ all real.

The greatest distance is from $-1 \pm i\sqrt3$ to $8:$ $\sqrt{(8 - (-1))^2 + (\sqrt3)^2}$ $= \sqrt{81 + 3}$ $= \sqrt{84} = 2\sqrt{21}.$

Thus, D is the correct answer.

16.

A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0),$ $(2020, 0),$ $(2020, 2020),$ and $(0, 2020).$ The probability that the point is within $d$ units of a lattice point is $\tfrac12.$ (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth?

$0.3$

$0.4$

$0.5$

$0.6$

$0.7$

Concepts:geometric probability circle area lattice point

Difficulty rating: 1730

Small Hint:

By periodicity, focus on a single unit square with a lattice point at each corner

Big Hint:

The four quarter-disks in one unit cell form one full disk of radius $d;$ set its area to $\tfrac12$

Solution:

By periodicity it suffices to consider one unit cell with a lattice point at each corner. The region within $d$ of a corner consists of four quarter-disks of radius $d,$ forming one full disk of area $\pi d^2.$ The resulting value will be less than $\tfrac12,$ so these quarter-disks do not overlap.

Setting $\pi d^2 = \tfrac12$ gives $d = \sqrt{\dfrac{1}{2\pi}} \approx 0.399.$

To the nearest tenth, $d = 0.4.$

Thus, B is the correct answer.

17.

The vertices of a quadrilateral lie on the graph of $y = \ln x,$ and the $x$ -coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln\dfrac{91}{90}.$ What is the $x$ -coordinate of the leftmost vertex?

$6$

$7$

$10$

$12$

$13$

Concepts:shoelace formula logarithm quadratic

Difficulty rating: 1860

Small Hint:

Let the $x$ -coordinates be $n, n+1, n+2, n+3$ and apply the shoelace formula

Big Hint:

The area simplifies to $\ln\dfrac{(n+1)(n+2)}{n(n+3)};$ set it equal to $\ln\dfrac{91}{90}$

Solution:

Let the vertices have $x$ -coordinates $n, n+1, n+2, n+3$ with $y$ -coordinates $\ln$ of those values. Applying the shoelace formula and simplifying, the area is $\ln\dfrac{(n+1)(n+2)}{n(n+3)}.$

Setting $\dfrac{(n+1)(n+2)}{n(n+3)} = \dfrac{91}{90}$ gives $1 + \dfrac{2}{n^2 + 3n} = \dfrac{91}{90},$ so $n^2 + 3n = 180.$

Then $(n - 12)(n + 15) = 0,$ so $n = 12.$

Thus, D is the correct answer.

18.

Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^\circ,$ $AC = 20,$ and $CD = 30.$ Diagonals $AC$ and $BD$ intersect at point $E,$ and $AE = 5.$ What is the area of quadrilateral $ABCD?$

$330$

$340$

$350$

$360$

$370$

Concepts:coordinate geometry circle area decomposition

Difficulty rating: 1800

Small Hint:

Place $A = (0,0),$ $C = (20, 0),$ and $D = (20, 30),$ so $E = (5, 0)$

Big Hint:

Since $\angle ABC = 90^\circ,$ $B$ lies on the circle with diameter $AC;$ it also lies on line $DE$

Solution:

Place $A = (0,0)$ and $C = (20, 0).$ Since $\angle ACD = 90^\circ,$ $D = (20, 30),$ and $E = (5, 0)$ because $AE = 5.$

Since $\angle ABC = 90^\circ,$ $B$ lies on the circle of radius $10$ centered at $(10, 0).$ Line $DE$ is $(5 + t,\, 2t);$ substituting gives $t^2 - 2t - 15 = 0,$ so $t = 5$ or $t = -3.$

For $E$ to lie between $B$ and $D,$ take $t = -3,$ giving $B = (2, -6),$ a distance $6$ below line $AC.$

Then $[ACD] = \tfrac12 \cdot 20 \cdot 30 = 300$ and $[ABC] = \tfrac12 \cdot 20 \cdot 6 = 60,$ so the total area is $360.$

Thus, D is the correct answer.

19.

There exists a unique strictly increasing sequence of nonnegative integers $a_1 \lt a_2 \lt \cdots \lt a_k$ such that $\frac{2^{289} + 1}{2^{17} + 1} = 2^{a_1} + 2^{a_2} + \cdots + 2^{a_k}.$ What is $k?$

$117$

$136$

$137$

$273$

$306$

Concepts:number base power of 2 factoring

Difficulty rating: 1990

Small Hint:

Write $x = 2^{17};$ then the quotient is $\dfrac{x^{17} + 1}{x + 1} = x^{16} - x^{15} + \cdots + 1$

Big Hint:

Group the alternating terms into pairs $x^{m+1} - x^m = 2^{17m}(2^{17} - 1),$ each a block of $17$ ones in binary

Solution:

Let $x = 2^{17}.$ Then $\dfrac{2^{289} + 1}{2^{17} + 1}$ $= \dfrac{x^{17} + 1}{x + 1}$ $= x^{16} - x^{15} + \cdots - x + 1,$ an alternating sum of the $17$ powers $x^0, x^1, \ldots, x^{16}.$

Pair each subtracted power with the added power just above it: $x^{m+1} - x^m$ $= 2^{17m}(2^{17} - 1)$ $= 2^{17m} + 2^{17m+1}$ $+ \cdots + 2^{17m+16},$ a block of $17$ consecutive powers of $2.$

There are $8$ such pairs, together with the leftover $+2^0.$ The blocks occupy disjoint ranges, so the total number of powers is $8 \cdot 17 + 1 = 137.$

Thus, C is the correct answer.

20.

Let $T$ be the triangle in the coordinate plane with vertices $(0, 0),$ $(4, 0),$ and $(0, 3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^\circ,$ $180^\circ,$ and $270^\circ$ counterclockwise around the origin, reflection across the $x$ -axis, and reflection across the $y$ -axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^\circ$ rotation, followed by a reflection across the $x$ -axis, followed by a reflection across the $y$ -axis will return $T$ to its original position, but a $90^\circ$ rotation, followed by a reflection across the $x$ -axis, followed by another reflection across the $x$ -axis will not return $T$ to its original position.)

$12$

$15$

$17$

$20$

$25$

Concepts:transformation casework

Difficulty rating: 1910

Small Hint:

Since $T$ is scalene, the only isometry fixing $T$ is the identity, so the three must compose to the identity

Big Hint:

The third transformation is forced by the first two; it is allowed exactly when the product of the first two is again one of the five maps

Solution:

Because $T$ is a scalene right triangle, the only isometry carrying $T$ to itself is the identity, so a sequence works exactly when the three transformations compose to the identity.

Let $r$ be the $90^\circ$ rotation and $s$ reflection across the $x$ -axis. The five allowed maps are $r,r^2,r^3,s,r^2s;$ the missing nonidentity maps are the diagonal reflections $rs,r^3s.$ In an ordered triple, the third map is forced by the first two and is allowed precisely when their product is one of the five.

Exactly $5$ ordered pairs have product identity: each first map is paired with its inverse. A diagonal reflection requires one rotation and one axis reflection; $r$ or $r^3$ may be paired on either side with $s$ or $r^2s,$ giving $2\cdot2\cdot2=8$ pairs. The remaining $25-5-8=12$ ordered pairs give valid sequences.

Thus, A is the correct answer.

21.

How many positive integers $n$ are there such that $n$ is a multiple of $5,$ and the least common multiple of $5!$ and $n$ equals $5$ times the greatest common divisor of $10!$ and $n?$

$12$

$24$

$36$

$48$

$72$

Concepts:least common multiple greatest common divisor prime factorization

Difficulty rating: 2080

Small Hint:

$5! = 2^3 \cdot 3 \cdot 5$ and $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7;$ compare exponents of each prime

Big Hint:

Multiplying the gcd by $5$ raises the power of $5$ by one; work out the allowed exponent of each prime in $n$

Solution:

Write $n = 2^a 3^b 5^c 7^d \cdots.$ Since $5! = 2^3 \cdot 3 \cdot 5$ has no other primes, $n$ can only involve $2, 3, 5, 7.$ Matching exponents in $\operatorname{lcm}(5!, n) = 5 \cdot \gcd(10!, n):$

For $2:$ $\max(3, a) = \min(8, a),$ so $3 \le a \le 8$ gives $6$ values. For $3:$ $\max(1, b) = \min(4, b),$ so $1 \le b \le 4$ gives $4$ values.

For $5:$ $\max(1, c) = 1 + \min(2, c)$ with $c \ge 1,$ which forces $c = 3,$ giving $1$ value. For $7:$ $\max(0, d) = \min(1, d),$ so $d = 0$ or $1,$ giving $2$ values.

The total is $6 \cdot 4 \cdot 1 \cdot 2 = 48.$

Thus, D is the correct answer.

22.

Let $(a_n)$ and $(b_n)$ be the sequences of real numbers such that $(2 + i)^n = a_n + b_n i$ for all integers $n \ge 0,$ where $i = \sqrt{-1}.$ What is $\sum_{n=0}^{\infty} \frac{a_n b_n}{7^n}?$

$\dfrac{3}{8}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

$\dfrac{9}{16}$

$\dfrac{4}{7}$

Concepts:complex number summation

Difficulty rating: 2110

Small Hint:

$a_n b_n$ $= \tfrac12 \operatorname{Im}\big((a_n + b_n i)^2\big)$ $= \tfrac12 \operatorname{Im}\big((2 + i)^{2n}\big)$

Big Hint:

Since $(2 + i)^2 = 3 + 4i,$ the sum becomes a geometric series with ratio $\dfrac{3 + 4i}{7}$

Solution:

Since $(a_n + b_n i)^2 = a_n^2 - b_n^2 + 2 a_n b_n i,$ we have $a_n b_n$ $= \tfrac12 \operatorname{Im}\big((2 + i)^{2n}\big)$ $= \tfrac12 \operatorname{Im}\big((3 + 4i)^n\big).$

Therefore the sum is $\tfrac12 \operatorname{Im} \displaystyle\sum_{n=0}^{\infty} \left(\frac{3 + 4i}{7}\right)^n$ $= \tfrac12 \operatorname{Im}\!\left(\frac{1}{1 - \frac{3 + 4i}{7}}\right).$

This equals $\tfrac12 \operatorname{Im}\!\left(\dfrac{7}{4 - 4i}\right)$ $= \tfrac12 \operatorname{Im}\!\left(\dfrac{7(4 + 4i)}{32}\right)$ $= \tfrac12 \cdot \dfrac{28}{32}$ $= \dfrac{7}{16}.$

Thus, B is the correct answer.

23.

Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly $7.$ Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?

$\dfrac{7}{36}$

$\dfrac{5}{24}$

$\dfrac{2}{9}$

$\dfrac{17}{72}$

$\dfrac{1}{4}$

Concepts:dice (probability)optimization casework

Difficulty rating: 2270

Small Hint:

Compare the win chance from rerolling one die (keep two summing to at most $6$ ) with rerolling two (keep one small die)

Big Hint:

Rerolling exactly two is best when no two dice sum to $6$ or less, yet the smallest die is small enough to beat rerolling all three

Solution:

Rerolling one die, keeping two dice that sum to $s,$ wins with probability $\tfrac16$ when $s \le 6$ and $0$ otherwise. Rerolling two dice, keeping a die of value $v,$ wins with probability equal to the number of ways two dice sum to $7 - v,$ over $36;$ this is largest when $v$ is smallest. For $v=1,2,3,$ these probabilities are $\tfrac5{36},\tfrac4{36},\tfrac3{36},$ all greater than the reroll-all probability $\tfrac{15}{216}=\tfrac{5}{72};$ for $v\ge4,$ rerolling all is better.

Rerolling exactly two dice is strictly best precisely when the two smallest dice sum to at least $7$ (so rerolling one cannot reach $7$ ) while the smallest die is $1, 2,$ or $3$ (so keeping it beats rerolling all three).

Sort the roll as $u\le v\le w.$ If $u=1,$ the only possibility is $(1,6,6),$ with $3$ orderings. If $u=2,$ the possibilities $(2,5,5),(2,5,6),(2,6,6)$ have $3+6+3=12$ orderings. If $u=3,$ choose $v,w$ with repetition from $\{4,5,6\};$ the six resulting triples have $3+6+6+3+6+3=27$ orderings. Thus there are $3+12+27=42$ qualifying ordered rolls out of $216,$ a probability of $\dfrac{42}{216} = \dfrac{7}{36}.$

Thus, A is the correct answer.

24.

Suppose that $\triangle ABC$ is an equilateral triangle of side length $s,$ with the property that there is a unique point $P$ inside the triangle such that $AP = 1,$ $BP = \sqrt{3},$ and $CP = 2.$ What is $s?$

$1 + \sqrt{2}$

$\sqrt{7}$

$\dfrac{8}{3}$

$\sqrt{5 + \sqrt{5}}$

$2\sqrt{2}$

Concepts:equilateral triangle quadratic

Difficulty rating: 2270

Small Hint:

For a point at distances $p, q, r$ from the vertices of an equilateral triangle of side $s,$ the four values satisfy $3(p^4 + q^4 + r^4 + s^4)$ $= (p^2 + q^2 + r^2 + s^2)^2$

Big Hint:

Substitute $p^2 = 1,$ $q^2 = 3,$ $r^2 = 4$ and solve the resulting quadratic in $s^2$

Solution:

A point at distances $p, q, r$ from the vertices of an equilateral triangle of side $s$ satisfies $3(p^4 + q^4 + r^4 + s^4)$ $= (p^2 + q^2 + r^2 + s^2)^2.$

With $p^2 = 1,$ $q^2 = 3,$ $r^2 = 4,$ letting $S = s^2$ gives $3(26 + S^2) = (8 + S)^2,$ so $S^2 - 8S + 7 = 0$ and $S = 1$ or $S = 7.$

A triangle of side $1$ cannot contain a point at distance $2$ from a vertex, so $S = 7$ and $s = \sqrt{7}.$

Thus, B is the correct answer.

25.

The number $a = \dfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying $\lfloor x \rfloor \cdot \{x\} = a \cdot x^2$ is $420,$ where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x.$ What is $p + q?$

$245$

$593$

$929$

$1331$

$1332$

Concepts:floor and ceiling functions quadratic summation

Difficulty rating: 2520

Small Hint:

On $x \in [n, n+1)$ write $\lfloor x \rfloor = n$ and $\{x\} = x - n,$ turning the equation into a quadratic $a x^2 - n x + n^2 = 0$

Big Hint:

Parameterize the two roots after dividing by $n$ as $1+\dfrac1u$ and $1+u;$ determine which positive integers $n$ put a root in $[n,n+1)$

Solution:

There are no negative solutions, while $x=0$ is always a solution. For $n\ge1$ and $x\in[n,n+1),$ put $y=\frac{x}{n}.$ The equation becomes $ay^2-y+1=0.$ Its roots must be real, so $0\lt a\le\tfrac14.$

If the two roots are $\alpha\le\beta,$ their sum and product are both $\frac{1}{a},$ so $(\alpha-1)(\beta-1)=1.$ Write $\alpha=1+\dfrac1u,\qquad \beta=1+u$ with $u\ge1.$ Then $a=\dfrac{u}{(u+1)^2}.$ The root $x=n\alpha$ lies in $[n,n+1)$ exactly when $n\lt u,$ while $n\beta$ never lies there for a positive integer $n.$

The required positive total ensures $u\gt1.$ Let $N$ be the largest positive integer less than $u,$ so $N\lt u\le N+1.$ The sum of all solutions is therefore $\dfrac{u+1}{u}\cdot\dfrac{N(N+1)}2=420.$ Because $\dfrac{u+1}{u}$ decreases with $u,$ these inequalities imply $\dfrac{N(N+2)}2\le420\lt\dfrac{(N+1)^2}{2},$ which forces $N=28.$

Substitution gives $406\cdot\dfrac{u+1}{u}=420,$ so $u=29.$ Hence $a=\dfrac{29}{30^2}=\dfrac{29}{900}.$ Indeed the positive solutions are $x=\dfrac{30n}{29}$ for $1\le n\le28,$ and their sum is $420.$ Therefore $p+q=29+900=929.$

Thus, C is the correct answer.