2023 AMC 10A 考试答案

1.

Cities $A$ and $B$ are $45$ miles apart. Alice and Beth start biking from $A$ and $B$ at speeds of $18$ mph and $12$ mph, respectively. How far away from city $A$ will they be when they meet?

$20$

$24$

$25$

$26$

$27$

Concepts:relative speed distance rate and time

Difficulty rating: 890

Solution:

They ride toward each other, so their speeds add. That closes the $45$ -mile gap at $18 + 12 = 30$ mph, and they meet after $45/30 = 1.5$ hours. Alice starts at $A,$ so by then she's gone $18 \cdot 1.5 = 27$ miles. Thus, E is the correct answer.

2.

The weight of $\frac{1}{3}$ of a large pizza together with $3\frac{1}{2}$ cups of orange slices is the same as the weight of $\frac{3}{4}$ of a large pizza together with $\frac{1}{2}$ cup of orange slices. A cup of orange slices weighs $\frac{1}{4}$ of a pound. What is the weight, in pounds, of a large pizza?

$1\frac{4}{5}$

$2$

$2\frac{2}{5}$

$3$

$3\frac{3}{5}$

Concepts:linear equation fraction

Difficulty rating: 1080

Solution:

Let $p$ be the pizza's weight. A cup of orange slices is $\frac14$ pound, so the two sides balance as $\frac13 p + \frac72 \cdot \frac14 = \frac34 p + \frac12 \cdot \frac14,$ that is $\frac13 p + \frac78 = \frac34 p + \frac18.$ Collect the pizza terms: $\frac68 = \left(\frac34 - \frac13\right)p = \frac{5}{12}p.$ So $p = \frac34 \cdot \frac{12}{5} = \frac95 = 1\frac45.$ Therefore, the answer is A.

3.

How many positive perfect squares less than $2023$ are divisible by $5?$

$8$

$9$

$10$

$11$

$12$

Concepts:perfect square divisibility

Difficulty rating: 1050

Solution:

If a perfect square is divisible by $5,$ it's divisible by $25,$ so it looks like $(5k)^2 = 25k^2.$ We need $25k^2 \lt 2023,$ i.e. $k^2 \lt 80.9.$ That allows $k = 1, 2, \ldots, 8,$ which is $8$ squares. Thus, A is the correct answer.

4.

A quadrilateral has all integer side lengths, a perimeter of $26,$ and one side of length $4.$ What is the greatest possible length of one side of this quadrilateral?

$9$

$10$

$11$

$12$

$13$

Concepts:triangle inequality bounding to limit cases

Difficulty rating: 1130

Solution:

In any quadrilateral each side is shorter than the sum of the other three. Call the longest side $s.$ The rest sum to $26 - s,$ so $s \lt 26 - s,$ which gives $s \lt 13$ and hence $s \le 12.$ Can we hit $12?$ The sides $4, 12, 9, 1$ work, since $12 \lt 4 + 9 + 1.$ So the greatest length is $12.$ Therefore, the answer is D.

5.

How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5?$

$14$

$15$

$16$

$17$

$18$

Concepts:prime factorization digits

Difficulty rating: 1200

Solution:

Factor everything into primes. $8^5 \cdot 5^{10} \cdot 15^5 = 2^{15} \cdot 5^{10} \cdot 3^5 \cdot 5^5 = 2^{15} \cdot 5^{15} \cdot 3^5 = 10^{15} \cdot 243.$ That's $243$ followed by $15$ zeros, so it has $3 + 15 = 18$ digits. Thus, E is the correct answer.

6.

An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21.$ What is the value of the cube?

$42$

$63$

$84$

$126$

$252$

Concepts:double counting cube geometry

Difficulty rating: 1270

Solution:

Count by incidences. Each edge lies on $2$ faces, so the six face values together are $2$ times the total of all edge values. Each vertex lies on $3$ edges, so the total edge value is $3$ times the vertex sum. Chaining these, the cube's value is $2 \cdot 3 \cdot 21 = 126.$ Therefore, the answer is D.

7.

Janet rolls a standard $6$ -sided die $4$ times and keeps a running total of the numbers she rolls. What is the probability that at some point her running total will equal $3?$

$\dfrac{2}{9}$

$\dfrac{49}{216}$

$\dfrac{25}{108}$

$\dfrac{17}{72}$

$\dfrac{13}{54}$

Concepts:dice (probability)casework

Difficulty rating: 1340

Solution:

The total can only reach exactly $3$ through the opening rolls, and these ways are disjoint: $3$ alone (probability $\frac16$ ), then $1,2$ and $2,1$ (each $\frac1{36}$ ), and $1,1,1$ (probability $\frac1{216}$ ). Add them up: $\frac{36}{216} + \frac{6}{216} + \frac{6}{216} + \frac{1}{216} = \frac{49}{216}.$ Thus, B is the correct answer.

8.

Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. Bread is done when its internal temperature is $200$ degrees Fahrenheit. What is this in degrees on the Breadus scale?

$33$

$34.5$

$36$

$37.5$

$39$

Concepts:linear equation slope

Difficulty rating: 1130

Solution:

The Breadus reading is linear in Fahrenheit through $(110, 0)$ and $(350, 100),$ so $B = \frac{100}{350 - 110}(F - 110) = \frac{5}{12}(F - 110).$ Plug in $F = 200:$ $B = \frac{5}{12} \cdot 90 = 37.5.$ Therefore, the answer is D.

9.

A digital display shows the current date as an $8$ -digit integer, consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For how many dates in $2023$ will each digit appear an even number of times in the digital display for that date?

$5$

$6$

$7$

$8$

$9$

Concepts:digits parity casework

Difficulty rating: 1410

Solution:

The year $2023$ already gives two $2$ s (even), one $0,$ and one $3.$ So to make every digit even overall, the four digits of $MM$ and $DD$ have to supply one more $0,$ one more $3,$ and then two digits equal to each other, all while keeping the count of $2$ s even. Run through the legal months and days of $2023$ under that rule and exactly $9$ dates survive. Thus, E is the correct answer.

10.

If Maureen scores an $11$ on her next quiz, her mean score will go up by $1.$ If she gets three $11$ s in a row, her mean score will increase by $2.$ What is her current mean quiz score?

$4$

$5$

$6$

$7$

$8$

Concepts:mean system of equations

Difficulty rating: 1270

Solution:

Let $m$ be the current mean over $n$ quizzes. One more $11$ makes the mean $m + 1:$ $\frac{mn + 11}{n + 1} = m + 1,$ which tidies up to $m + n = 10.$ Three more $11$ s make it $m + 2:$ $\frac{mn + 33}{n + 3} = m + 2,$ i.e. $3m + 2n = 27.$ Solve the pair and $m = 7.$ Therefore, the answer is D.

11.

A square with area $3$ has a square with area $2$ inscribed in it. This creates $4$ smaller congruent right triangles. What is the ratio of the smaller leg to the larger leg in the shaded right triangle?

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$2 - \sqrt{3}$

$\sqrt{3} - \sqrt{2}$

$\sqrt{2} - 1$

Concepts:square (geometry)Pythagorean Theorem Vieta’s Formulas

Difficulty rating: 1500

Solution:

Each corner right triangle has legs $a$ and $b.$ A side of the outer square gives $a+b=\sqrt3,$ and a side of the inscribed square gives $a^2+b^2=2.$ Subtract to find the product: $2ab=(a+b)^2-(a^2+b^2)=3-2=1,$ so $ab=\tfrac12.$ Then $a$ and $b$ are the roots of $t^2-\sqrt3\,t+\tfrac12=0,$ namely $\tfrac{\sqrt3\pm1}{2}.$ The ratio of the smaller leg to the larger is $\dfrac{\sqrt3-1}{\sqrt3+1}=\dfrac{(\sqrt3-1)^2}{2}=2-\sqrt3.$ Thus, C is the correct answer.

12.

How many three-digit positive integers $N$ satisfy both of the following properties: $N$ is divisible by $7,$ and the number formed by reversing the digits of $N$ is divisible by $5?$

$13$

$14$

$15$

$16$

$17$

Concepts:divisibility digits counting integers in a range

Difficulty rating: 1440

Solution:

When we reverse $N,$ its last digit is the first digit of $N.$ For the reversal to be divisible by $5,$ that digit is $0$ or $5.$ A three-digit number can't start with $0,$ so $N$ starts with $5,$ meaning $500 \le N \le 599$ (and the reversal ends in $5,$ always fine). Now just count multiples of $7$ here: from $7 \cdot 72 = 504$ to $7 \cdot 85 = 595,$ that's $14$ numbers. Therefore, the answer is B.

13.

Abdul and Chiang are standing $48$ feet apart in a field. Bharat is standing in the same field as far from Abdul as possible so that the angle formed by his lines of sight to Abdul and Chiang measures $60^\circ.$ What is the square of the distance (in feet) between Abdul and Bharat?

$1728$

$2601$

$3072$

$4608$

$6912$

Concepts:inscribed angle law of sines optimization

Difficulty rating: 1590

Solution:

Let $A$ be Abdul, $C$ be Chiang with $AC = 48,$ and $B$ be Bharat with $\angle B = 60^\circ.$ Every point seeing $AC$ at $60^\circ$ lies on one circular arc, so all valid $B$ sit on a circle where chord $AC$ subtends $60^\circ.$ The law of sines gives its diameter, $\frac{AC}{\sin 60^\circ} = \frac{48}{\sqrt3/2} = 32\sqrt3.$ Now $AB$ is a chord, and a chord is longest when it's a diameter. So $AB = 32\sqrt3$ and $AB^2 = 1024 \cdot 3 = 3072.$ Thus, C is the correct answer.

14.

A number is chosen at random from among the first $100$ positive integers, and a positive integer divisor of that number is then chosen at random. What is the probability that the chosen divisor is divisible by $11?$

$\dfrac{4}{100}$

$\dfrac{9}{200}$

$\dfrac{1}{20}$

$\dfrac{11}{200}$

$\dfrac{3}{50}$

Concepts:factor counting basic probability

Difficulty rating: 1630

Solution:

A number $n \le 100$ can only have a divisor divisible by $11$ when $11 \mid n,$ so $n \in \{11, 22, \ldots, 99\}.$ Write $n = 11m$ with $m \le 9.$ Here $11 \nmid m,$ so $d(11m) = 2\,d(m),$ and the divisors that are multiples of $11$ are exactly the $d(m)$ numbers $11d.$ That makes the chance $\frac{d(m)}{2\,d(m)} = \frac12$ for each such $n.$ Averaging over all $100$ starting numbers, the probability is $\frac{1}{100}\sum_{m=1}^{9}\frac12 = \frac{9}{200}.$ Therefore, the answer is B.

15.

An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi?$

$46$

$48$

$56$

$60$

$64$

Concepts:circle area arithmetic sequence summation

Difficulty rating: 1560

Solution:

A circle of radius $r$ has area $\pi r^2.$ So the shaded ring between radius $2k$ and $2k-1$ has area $\pi\big((2k)^2 - (2k-1)^2\big) = (4k-1)\pi.$ With $2n$ circles the shaded total is $\pi\sum_{k=1}^{n}(4k-1) = \pi(2n^2 + n).$ We want $2n^2 + n \ge 2023.$ At $n = 31$ it's $1953,$ at $n = 32$ it's $2080.$ So $n = 32,$ which means $2n = 64$ circles. Thus, E is the correct answer.

16.

In a tennis tournament, each person plays every other person once. In this tournament, there are twice as many right-handed players as left-handed players, but left-handed players won $40\%$ more games than right-handed players. How many total games were played?

$15$

$36$

$45$

$48$

$66$

Concepts:combinations ratio and proportion

Difficulty rating: 1730

Solution:

Say there are $L$ left-handers and $2L$ right-handers, so $3L$ players and $\binom{3L}{2}$ games. Every game has one winner, and left wins are $1.4$ times right wins, so the wins split $7 : 5$ and the total must be a multiple of $12.$ Try $L = 3:$ that's $\binom{9}{2} = 36 = 12 \cdot 3,$ with left winning $21$ and right winning $15,$ and indeed $21 = 1.4 \cdot 15.$ It's achievable. So $36$ games were played. Therefore, the answer is B.

17.

Let $ABCD$ be a rectangle with $AB = 30$ and $BC = 28.$ Points $P$ and $Q$ lie on $BC$ and $CD$ respectively so that all sides of $\triangle ABP,$ $\triangle PCQ,$ and $\triangle QDA$ have integer lengths. What is the perimeter of $\triangle APQ?$

$84$

$86$

$88$

$90$

$92$

Concepts:Pythagorean Triple rectangle systematic listing

Difficulty rating: 1840

Solution:

Set $A = (0,0),$ $B = (30,0),$ $C = (30,28),$ $D = (0,28),$ with $P = (30, p)$ on $BC$ and $Q = (30 - q, 28)$ on $CD.$ The three right triangles give $AP = \sqrt{30^2 + p^2},$ $QA = \sqrt{28^2 + (30 - q)^2},$ and $PQ = \sqrt{(28 - p)^2 + q^2},$ and all must be integers. Hunt for Pythagorean triples: $p = 16$ makes $AP = 34,$ and $30 - q = 21$ (so $q = 9$ ) makes $QA = 35.$ Then $PQ = \sqrt{12^2 + 9^2} = 15,$ an integer too. So the perimeter of $\triangle APQ$ is $34 + 15 + 35 = 84.$ Thus, A is the correct answer.

18.

A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meeting?

$5$

$6$

$7$

$8$

$9$

Concepts:Euler’s Polyhedron Formula polyhedron graph theory double counting

Difficulty rating: 1660

Solution:

Each rhombus has $4$ edges, and every edge is shared by $2$ faces, so $E = \frac{12 \cdot 4}{2} = 24.$ With $F = 12,$ Euler's formula gives $V = 2 - F + E = 14.$ Suppose $x$ vertices have $3$ edges and the other $14 - x$ have $4.$ The degrees sum to twice the edge count: $3x + 4(14 - x) = 2E = 48,$ so $x = 8.$ Therefore, the answer is D.

19.

The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s).$ What is $|r - s|?$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$\dfrac{3}{4}$

$\dfrac{2}{3}$

$1$

Concepts:transformation perpendicular bisector coordinate geometry

Difficulty rating: 1730

Solution:

A rotation keeps its center equidistant from each point and its image. So $P$ is equidistant from $A$ and $A',$ and from $B$ and $B',$ which puts it at the intersection of two perpendicular bisectors. The bisector of $BB'$ from $(3,3)$ to $(4,3)$ is $x = 3.5.$ The bisector of $AA'$ from $(1,2)$ to $(3,1)$ is $2x - y = 2.5.$ Then $y = 2(3.5) - 2.5 = 4.5,$ so $P = (3.5, 4.5)$ and $|r - s| = |3.5 - 4.5| = 1.$ Thus, E is the correct answer.

20.

Each square in a $3 \times 3$ grid of squares is colored red, white, blue, or green so that every $2 \times 2$ square contains one square of each color. One such coloring is shown below (letters denote the colors, with the center square white). How many different colorings are possible?

$24$

$48$

$60$

$72$

$96$

Concepts:arrangements with restrictions casework

Difficulty rating: 2080

Solution:

Label the cells row by row $a, b, c / d, e, f / g, h, i.$ The top-left block $a, b, d, e$ is a permutation of the four colors, so $4! = 24$ ways. The block $\{b, c, e, f\}$ is also all four colors, and $b, e$ are fixed, so $\{c, f\}$ is the remaining two in some order: $2$ ways. Same story for $\{g, h\},$ the two colors apart from $d, e,$ another $2$ ways. That leaves $i,$ forced to whatever color is missing from $\{e, f, h\},$ and that only works when $f \ne h.$ Of the $2 \cdot 2 = 4$ order combinations, exactly one has $f = h,$ so $3$ survive. The total is $24 \cdot 3 = 72.$ Therefore, the answer is D.

21.

There is a unique polynomial $P(x)$ of least degree with leading coefficient $1$ satisfying all of the following:

$1$ is a root of $P(x) - 1,$ $2$ is a root of $P(x - 2),$ $3$ is a root of $P(3x),$ and $4$ is a root of $4P(x).$

All the roots of $P(x)$ except one are integers. If the one non-integer root can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, what is $m + n?$

$41$

$43$

$45$

$47$

$49$

Concepts:polynomial substitution

Difficulty rating: 2120

Solution:

Translate each condition into a value: $P(1) = 1,$ $P(0) = 0,$ $P(9) = 0,$ and $P(4) = 0.$ So $0, 4, 9$ are roots. Could a cubic do it? A monic cubic with those roots has $P(1) = (1)(-3)(-8) = 24 \ne 1,$ so no. The least-degree monic polynomial is degree $4:$ $P(x) = x(x - 4)(x - 9)(x - c).$ Now $P(1) = (1)(-3)(-8)(1 - c) = 24(1 - c) = 1,$ so $1 - c = \frac{1}{24}$ and $c = \frac{23}{24}.$ That's the lone non-integer root, so $m + n = 23 + 24 = 47.$ Thus, D is the correct answer.

22.

Circles $C_1$ and $C_2$ have radius $1,$ and the distance between their centers is $\frac{1}{2}.$ Circle $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2.$ Circle $C_4$ is internally tangent to both $C_1$ and $C_2$ and is externally tangent to $C_3.$ What is the radius of $C_4?$

$\dfrac{1}{14}$

$\dfrac{1}{12}$

$\dfrac{1}{10}$

$\dfrac{3}{28}$

$\dfrac{1}{9}$

Concepts:tangent circles coordinate geometry symmetry

Difficulty rating: 2270

Solution:

Put the centers of $C_1, C_2$ at $\left(\pm\frac14, 0\right).$ By symmetry the largest circle inside both sits at the origin with radius $r_3,$ where $1 - r_3 = \frac14,$ so $r_3 = \frac34.$ Let $C_4$ be centered at $(0, y)$ with radius $r.$ Internal tangency to $C_1$ gives $\sqrt{\frac1{16} + y^2} = 1 - r,$ and external tangency to $C_3$ gives $y = \frac34 + r.$ Substitute the second into the first: $\frac1{16} + \left(\frac34 + r\right)^2 = (1 - r)^2.$ This collapses to $\frac72 r = \frac38,$ so $r = \frac{3}{28}.$ Therefore, the answer is D.

23.

Positive integer divisors $a$ and $b$ of $N$ are called complementary if $ab = N.$ Given that $N$ has a pair of complementary divisors that differ by $20$ and a pair of complementary divisors that differ by $23,$ find the sum of the digits of $N.$

$11$

$13$

$15$

$17$

$19$

Concepts:difference of squares factor Diophantine Equation

Difficulty rating: 2380

Solution:

Complementary divisors differing by $20$ are $b$ and $b + 20$ with product $N,$ so $N = b^2 + 20b$ and $N + 100 = (b + 10)^2.$ A pair differing by $23$ gives $4N + 529 = (2d + 23)^2.$ Set $N + 100 = k^2.$ Then $4k^2 + 129 = m^2,$ so $(m - 2k)(m + 2k) = 129 = 3 \cdot 43.$ Take the factorization $1 \cdot 129:$ it gives $m = 65,$ $k = 32,$ hence $N = 32^2 - 100 = 924.$ Check it: $924 = 22 \cdot 42 = 21 \cdot 44,$ and the digit sum is $9 + 2 + 4 = 15.$ Thus, C is the correct answer.

24.

Six regular hexagonal blocks of side length $1$ unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is $\frac{3}{7}$ unit. What is the area of the region inside the frame not occupied by the blocks?

$\dfrac{13\sqrt{3}}{3}$

$\dfrac{216\sqrt{3}}{49}$

$\dfrac{9\sqrt{3}}{2}$

$\dfrac{14\sqrt{3}}{3}$

$\dfrac{243\sqrt{3}}{49}$

Concepts:regular polygon area area decomposition

Difficulty rating: 2520

Solution:

The uncovered region is the frame's area minus the six unit blocks. A regular hexagon of side $t$ has area $\tfrac{3\sqrt3}{2}t^2,$ so each unit block is $\tfrac{3\sqrt3}{2}.$ The spacing rule, that each frame corner sits $\tfrac37$ from the nearest block vertex, pins the frame's side length at $3.$ So the uncovered area is $\tfrac{3\sqrt3}{2}\cdot 3^2 - 6\cdot\tfrac{3\sqrt3}{2} = \tfrac{27\sqrt3}{2} - 9\sqrt3 = \tfrac{9\sqrt{3}}{2}.$ Therefore, the answer is C.

25.

If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A, B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B.$ For example, if $AB$ is an edge of the polyhedron, then $d(A, B) = 1,$ but if $AC$ and $CB$ are edges and $AB$ is not an edge, then $d(A, B) = 2.$ Let $Q,$ $R,$ and $S$ be randomly chosen distinct vertices of a regular icosahedron (a regular polyhedron made up of $20$ equilateral triangles). What is the probability that $d(Q, R) \gt d(R, S)?$

$\dfrac{7}{22}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

$\dfrac{5}{12}$

$\dfrac{1}{2}$

Concepts:graph theory symmetry basic probability

Difficulty rating: 2600

Solution:

Fix $R.$ Of the other $11$ vertices, $5$ sit at distance $1,$ $5$ at distance $2,$ and $1$ (the opposite vertex) at distance $3.$ Pick ordered distinct $Q, S$ from these $11:$ that's $11 \cdot 10 = 110$ pairs. The ones with $d(R,Q) = d(R,S)$ number $5\cdot4 + 5\cdot4 + 1\cdot0 = 40,$ so $P(\text{equal}) = \frac{40}{110} = \frac{4}{11}.$ By the symmetry between $Q$ and $S,$ the $\gt$ and $\lt$ cases split the rest evenly, so $P(d(Q,R) \gt d(R,S)) = \frac{1 - \frac4{11}}{2} = \frac{7}{22}.$ Thus, A is the correct answer.