AMC 10A 2024 Problems & Solutions | Step-by-Step Math Hints

Find all AMC 10A 2024 problems with step-by-step hints, detailed solutions, and final answers — perfect for students preparing for AMC 10 or AIME.

Problem 1

What is the value of $9901\cdot101 - 99\cdot10101$ ?

2
20
200
202
2020

Hints & Final Answer

Hint 1

(9900+1)(100+1)-(99)(10000+100+1)

Hint 2

\cancel{990000} + \bcancel{9900} + 100 + 1 - \cancel{990000} - \bcancel{9900} - 99

Final Answer

(A) 2

Problem 2

A model used to estimate the time it will take to hike to the top of the mountain on a trail is of the form $T = aL + bG$ , where $a$ and $b$ are constants, $T$ is the time in minutes, $L$ is the length of the trail in miles, and $G$ is the altitude gain in feet. The model estimates that it will take 69 minutes to hike to the top if a trail is 1.5 miles long and ascends 800 feet, as well as if a trail is 1.2 miles long and ascends 1100 feet. How many minutes does the model estimate it will take to hike to the top if the trail is 4.2 miles long and ascends 4000 feet?

Hints & Final Answer

Hint 1

69 = a(1.5) + b(800)

Hint 2

69 = a(1.2) + b(1100)

Hint 3

Find

a

and

b

Hint 4

a = 30, \quad b = 0.03

Hint 5

T = aL + bG = 30(4.2) + 0.03(4000) = 126 + 120

Final Answer

(B) 246

Problem 3

What is the sum of the digits of the smallest prime that can be written as a sum of 5 distinct primes?

Hints & Final Answer

Hint 1

The sum of five odd numbers is odd.

Hint 2

The number

2

is not among the five distinct primes.

Hint 3

3 + 5 + 7 + 11 + 13 = 39

Hint 4

Try increasing the largest prime.

Hint 5

3 + 5 + 7 + 11 + 17 = 43

Final Answer

(B) 7

Problem 4

The number 2024 is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?

Hints & Final Answer

Hint 1

Use as many 99s as possible:

2024 = 99 + \cdots + 99 + ?

Hint 2

Divide 2024 by 99.

Hint 3

2024 = 99\cdot 20 + 44.

Hint 4

2024 = \underbrace{99+99+\cdots+99}_{20\ \text{terms}} + 44

Final Answer

(B) 21

Problem 5

What is the least value of n such that $n!$ is a multiple of 2024?

Hints & Final Answer

Hint 1

Prime factorize 2024.

Hint 2

2024 = 2^3 \times 11 \times 23

Hint 3

n!

should be a multiple of 23.

Final Answer

(D) 23

Problem 6

What is the minimum number of successive swaps of adjacent letters in the string $ABCDEF$ that are needed to change the string to $FEDCBA$ ? (For example, 3 swaps are required to change $ABC$ to $CBA$ ; one such sequence of swaps is $ABC \rightarrow BAC \rightarrow BCA \rightarrow CBA.$ )

Hints & Final Answer

Hint 1

ABCDEF

, we can move

F

to the first position with 5 swaps.

Hint 2

ABCDEF \xrightarrow{5} FABCDE \xrightarrow{4} FEABCD \xrightarrow{3} FEDABC \xrightarrow{2} FEDCAB \xrightarrow{1} FEDCBA

Hint 3

The answer

\ge 15.

Hint 4

If you want to prove rigorously that the answer is not less than 15, you can assign 1 to 6 to

A

through

F

, respectively, and use the number of inversions

(i < j, a_i > a_j).

Final Answer

(D) 15

Problem 7

The product of three integers is 60. What is the least possible positive sum of the three integers?

Hints & Final Answer

Hint 1

abc = 60 = 2 \times 2 \times 3 \times 5

Hint 2

60 = 10 \times (-6) \times (-1)

Hint 3

\text{Ans} \leq 10 - 6 - 1 = 3

Hint 4

If you check casework, you can prove that the answer ≥ 3.

Final Answer

(B) 3

Problem 8

Amy, Bomani, Charlie, and Daria work in a chocolate factory. On Monday Amy, Bomani, and Charlie started working at 1:00 PM and were able to pack 4, 3, and 3 packages, respectively, every 3 minutes. At some later time, Daria joined the group, and Daria was able to pack 5 packages every 4 minutes. Together, they finished packing 450 packages at exactly 2:45 PM. At what time did Daria join the group?

1:25 PM
1:35 PM
1:45 PM
1:55 PM
2:05 PM

Hints & Final Answer

Hint 1

Calculate the number of packages that Amy, Bomani, Charlie, and Daria can complete per minute.

Hint 2

Amy, Bomani, Charlie, and Daria can complete

\frac{4}{3},\,1,\,1,\,\text{and } \frac{5}{4}

packages per minute respectively.

Hint 3

Assume Daria joined

t

minutes after 1 PM. Write an equation in terms of

t

Hint 4

From 1 PM to 2:45 PM is 105 minutes. Timeline: 1 PM ——— t minutes (A, B, C) ——— 105 − t minutes (A, B, C, D) ——— 2:45 PM.

Hint 5

t\!\left(\tfrac{4}{3}+1+1\right) + (105 - t)\!\left(\tfrac{4}{3}+1+1+\tfrac{5}{4}\right) = 450

Final Answer

(A) 1:25 PM

Problem 9

In how many ways can 6 juniors and 6 seniors form 3 disjoint teams of 4 people so that each team has 2 juniors and 2 seniors?

720
1350
2700
3280
8100

Hints & Final Answer

Hint 1

The first team has

\binom{6}{2}\binom{6}{2}

cases.

Hint 2

If the teams had names, for example, Team1, Team2, and Team3, then there would be

\binom{6}{2}^2 \binom{4}{2}^2 \binom{2}{2}^2

cases.

Hint 3

Since the teams do not have names, you should divide the result in Hint 2 by

3!

, since we are counting the permutations of the teams.

Final Answer

(B) 1350

Problem 10

Consider the following operation. Given a positive integer n, if n is a multiple of 3, then you replace n by $\frac{n}{3}$ . If n is not a multiple of 3, then you replace n by $n + 10$ . For example, beginning with $n = 4$ , this procedure gives $4 \rightarrow 14 \rightarrow 24 \rightarrow 8 \rightarrow 18 \rightarrow 6 \rightarrow 2 \rightarrow 12 \rightarrow \cdots$ . Suppose you start with $n = 100$ . What value results if you perform this operation exactly 100 times?

Hints & Final Answer

Hint 1

Write the first few steps to find the pattern.

Hint 2

100 \rightarrow 110 \rightarrow 120 \rightarrow 40 \rightarrow 50 \rightarrow 60 \rightarrow 20 \rightarrow 30 \rightarrow 10 \rightarrow 20 \rightarrow 30 \rightarrow 10 \rightarrow \cdots

Hint 3

100 → 110 → 120 → 40 → 50 → 60 → 20 → 30 → 10

(20 → 30 → 10 repeat in a loop)

Hint 4

After 6 steps, there is a loop with a length of 3.

Hint 5

(100 - 6) \equiv 1 \pmod{3}

Final Answer

Problem 11

How many ordered pairs of integers (m, n) satisfy $\sqrt{n^2-49}=m$ ?

1
2
3
4
infinitely many

Hints & Final Answer

Hint 1

Square both sides.

Hint 2

n^2-49=m^2

Hint 3

Use the difference of two squares identity.

Hint 4

(n-m)(n+m)=49

Hint 5

\sqrt{n^2-49}=m \;\Rightarrow\; m\ge 0 \;\Rightarrow\; n-m \le n+m

Hint 6

n−m and n+m are factors of 49, where n−m ≤ n+m.

Hint 7

(n-m,\,n+m)=(1,49),(7,7),(-7,-7),(-49,-1)

Hint 8

(n,\,m)=(25,24),(7,0),(-7,0),(-25,24)

Final Answer

(D) 4

Problem 12

Zelda played the Adventures of Math game on August 1 and scored 1,700 points. She continued to play daily over the next 5 days. The bar chart below shows the daily change in her score compared to the day before. (For example, Zelda’s score on August 2 was 1,700 + 80 = 1,780 points.) What was Zelda’s average score in points over the 6 days?

1700
1702
1703
1713
1715

Hints & Final Answer

Hint 1

S_1 = 1700,\quad S_2 = 1700 + 80 = 1780,\quad S_3 = 1780 - 90 = 1690

Hint 2

S_4 = 1690 - 10 = 1680,\quad S_5 = 1680 + 60 = 1740,\quad S_6 = 1740 - 40 = 1700

Hint 3

\text{Average} = \dfrac{S_1 + S_2 + S_3 + S_4 + S_5 + S_6}{6} = 1700 + \dfrac{0 + 80 - 90 - 10 + 60 - 40}{6} = 1715

Final Answer

(E) 1715

Problem 13

Two transformations are said to commute if applying the first followed by the second gives the same result as applying the second followed by the first. Consider these four transformations of the coordinate plane:

a translation 2 units to the right,
a 90°-rotation counterclockwise about the origin,
a reflection across the x-axis, and
a dilation centered at the origin with scale factor 2.

Of the 6 pairs of distinct transformations from this list, how many commute?

Hints & Final Answer

Hint 1

Transformation 1:

(x,y)\mapsto(x+2,\;y)

Hint 2

Transformation 2:

(x,y)\mapsto(-y,\;x)

Hint 3

Transformation 3:

(x,y)\mapsto(x,\;-y)

Hint 4

Transformation 4:

(x,y)\mapsto(2x,\;2y)

Hint 5

For

T_1

and

T_2

(x,y)\xrightarrow{T_1}(x+2,y)\xrightarrow{T_2}(-y,\;x+2)\\ (x,y)\xrightarrow{T_2}(-y,x)\xrightarrow{T_1}(-y+2,\;x)

T_1

and

T_2

do not commute.

Hint 6

For

T_1

and

T_3

(x,y)\xrightarrow{T_1}(x+2,y)\xrightarrow{T_3}(x+2,-y)\\ (x,y)\xrightarrow{T_3}(x,-y)\xrightarrow{T_1}(x+2,-y)

T_1

and

T_3

commute.

Hint 7

For

T_1

and

T_4

(x,y)\xrightarrow{T_1}(x+2,y)\xrightarrow{T_4}(2x+4,\,2y)\\ (x,y)\xrightarrow{T_4}(2x,2y)\xrightarrow{T_1}(2x+2,\,2y)

T_1

and

T_4

do not commute.

Hint 8

For

T_2

and

T_3

(x,y)\xrightarrow{T_2}(-y,x)\xrightarrow{T_3}(-y,-x)\\ (x,y)\xrightarrow{T_3}(x,-y)\xrightarrow{T_2}(y,x)

T_2

and

T_3

do not commute.

Hint 9

For

T_2

and

T_4

(x,y)\xrightarrow{T_2}(-y,x)\xrightarrow{T_4}(-2y,\,2x)\\ (x,y)\xrightarrow{T_4}(2x,2y)\xrightarrow{T_2}(-2y,\,2x)

T_2

and

T_4

commute.

Hint 10

For

T_3

and

T_4

(x,y)\xrightarrow{T_3}(x,-y)\xrightarrow{T_4}(2x,-2y)\\ (x,y)\xrightarrow{T_4}(2x,2y)\xrightarrow{T_3}(2x,-2y)

T_3

and

T_4

commute.

Final Answer

Problem 14

One side of an equilateral triangle of height 24 lies on line ℓ. A circle of radius 12 is tangent to line ℓ and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line ℓ can be written as $a\sqrt{b}-c\pi$ , where $a,b,c$ are positive integers and $b$ is not divisible by the square of any prime. What is $a+b+c$ ?

Hints & Final Answer

Hint 1

Hint 2

Hint 3

Hint 4

The kite-like region can be decomposed into two congruent 30–60–90 triangles.

[ODBE] \;=\; 2\cdot \frac{12\cdot \tfrac{12}{\sqrt{3}}}{2} \;=\; 48\sqrt{3}.

Hint 5

The circular piece cut out is a 60° sector of radius 12.

[\text{Sector }OED] \;=\; \frac{60}{360}\,\pi r^2 \;=\; \frac{1}{6}\pi (12)^2 \;=\; 24\pi.

Hint 6

Subtract the sector from the two-triangle area.

\text{Desired area} \;=\; 48\sqrt{3}-24\pi.

Final Answer

(D) 48 + 3 + 24 = 75

Problem 15

Let $M$ be the greatest integer such that both $M + 1213$ and $M + 3773$ are perfect squares. What is the units digit of $M$ ?

Hints & Final Answer

Hint 1

Let

M + 1213 = x^2

and

M + 3773 = y^2.

Hint 2

Without loss of generality, assume

x, y > 0.

Hint 3

y^2 - x^2 = 2560.

Hint 4

Use the identity of the difference of two squares.

Hint 5

(y - x)(y + x) = 2560.

Hint 6

To maximize

M

, we should maximize

y,

since

M + 3773 = y^2.

Hint 7

From Hint 5, both

y - x

and

y + x

must be even.

Hint 8

y - x = 2,

y + x = 1280.

Hint 9

y = 641.

Hint 10

M = y^2 - 3773 = 641^2 - 3773.

Hint 11

641^2 - 3773 \equiv 1^2 - 3 \equiv -2 \equiv 8 \pmod{10}.

Final Answer

(E) 8

Problem 16

All of the rectangles in the figure below, which is drawn to scale, are similar to the enclosing rectangle. Each number represents the area of the rectangle. What is length AB?

Rectangles similar to the enclosing rectangle

$4 + 4\sqrt{5}$
$10\sqrt{2}$
$5 + 5\sqrt{5}$
$10\sqrt{8}$
$20$

Hints & Final Answer

Hint 1

Let the smaller and greater sides of the rectangle with area 1 be

x

and

y

, respectively.

Hint 2

Hint 3

CD=\sqrt{18}\,y=3\sqrt{2}\,y.

Hint 4

CD = \overline{BD}-x = 7x - x = 6x.

Hint 5

xy=1,\qquad 3\sqrt{2}\,y = 6x.

Hint 6

y = x\sqrt{2},\quad x=\dfrac{1}{4\sqrt{2}}.

Hint 7

AB = 4\sqrt{2}\,y + 5x + 7x.

Hint 8

AB = 4\sqrt{2}\,(x\sqrt{2}) + 12x = 20x.

Final Answer

(D) 10\sqrt{8}

Problem 17

Two teams are in a best-two-out-of-three playoff: the teams will play at most 3 games, and the winner of the playoff is the first team to win 2 games. The first game is played on Team A’s home field, and the remaining games are played on Team B’s home field. Team A has a $\tfrac{2}{3}$ chance of winning at home, and its probability of winning when playing away from home is $p$ . Outcomes of the games are independent. The probability that Team A wins the playoff is $\tfrac{1}{2}$ . Then $p$ can be written in the form $\tfrac12 (m - \sqrt{n})$ , where $m$ and $n$ are positive integers. What is $m + n$ ?

Hints & Final Answer

Hint 1

There are four cases.

Hint 2

Case       A's home field     B's home field     B's home field
1                wins             wins
2                wins             loses             wins
3                loses            wins              wins

Hint 3

Case 1:

\tfrac{2}{3}p

Hint 4

Case 2:

\tfrac{2}{3}(1-p)p

Hint 5

Case 3:

\tfrac{1}{3}p^2

Hint 6

Write an equation in terms of p

Hint 7

\tfrac{2}{3}p + \tfrac{2}{3}(1-p)p + \tfrac{1}{3}p^2 = \tfrac{1}{2}

Hint 8

2p^2 - 8p + 3 = 0

Hint 9

Use the quadratic formula

Hint 10

p = \frac{8 \pm \sqrt{64 - 24}}{4}

Hint 11

p = \frac{4 \pm \sqrt{10}}{2}

Hint 12

0 \le p \le 1

Hint 13

p = \frac{4 - \sqrt{10}}{2}

Final Answer

(E)

4 + 10 = 14

Problem 18

There are exactly K positive integers 5 ≤ b ≤ 2024 such that the base-b integer 2024_b is divisible by 16 (where 16 is in base ten). What is the sum of the digits of K?

Hints & Final Answer

Hint 1

2024_b = 4 + 2b + 0b² + 2b³

Hint 2

2b³ + 2b + 4 ≡ 0 (mod 16)

Hint 3

b³ + b + 2 ≡ 0 (mod 8)

Hint 4

b ≡ 0, 1, −1, 2, −2, 3, −3, 4 (mod 8)

Hint 5

b³ + b + 2 ≡ 2, 4, 0, 4, 0, 0, 4, 6 (mod 8)

Hint 6

→ b ≡ −1, −2, 3 ≡ 7, 6, 3 (mod 8)

Hint 7

b ≡ 3 (mod 8) → = 2024 8 − 1 = 253 − 1 = 252

Hint 8

b ≡ 6 (mod 8) → = 2024 8 = 253

Hint 9

b ≡ 7 (mod 8) → = 2024 8 = 253

Hint 10

252 + 253 + 253 = 758

Final Answer

(D) 7 + 5 + 8 = 20

Problem 19

The first three terms of a geometric sequence are the integers a, 720 and b, where a < 720 < b. What is the sum of the digits of the least possible value of b?

Hints & Final Answer

Hint 1

Common Ratio :

\dfrac{720}{a} = \dfrac{b}{720} \Rightarrow ab = 720^2

Hint 2

ab = 2^8 3^4 5^2

Hint 3

To rigorously find the minimum

b

and prove that it is indeed the minimum, you can use casework.

Hint 4

Use casework based on the power of 2 in

b

Hint 5

Case 1:

b = 2^8 \times \cdots = 256 \times \cdots > 720

\Rightarrow \min\{b\} = 256 \times 3 = 768

Hint 6

Case 2:

b = 2^7 \times \cdots = 128 \times \cdots > 720

\Rightarrow \min\{b\} = 128 \times 9 = 1152 > 768 \text{ (from Hint 5)}

Hint 7

For the remaining cases, if we write down the positive divisors of

3^4 5^2

first, then it will be faster to find the minimum

b

in each case.

Hint 8

Positive divisors of

3^4 5^2

\{1,3,5,9,15,25,27,45,75,81,135,225,405,675,2025\}

Hint 9

b = 64\times 15 > 768

✗

b = 32\times 25 = 800 > 768

✗

b = 16\times 75 > 768

✗

b = 8\times 135 > 768

✗

b = 4\times 225 > 768

✗

b = 2\times 405 > 768

✗

b = 1\times 2025 > 768

✗

Final Answer

(E)

7+6+8=21

Problem 20

Let S be a subset of {1, 2, 3, …, 2024} such that the following two conditions hold: • If x and y are distinct elements of S, then |x − y| > 2. • If x and y are distinct odd elements of S, then |x − y| > 6. What is the maximum possible number of elements in S?

Hints & Final Answer

Hint 1

1 +3 → 4 +4 → 8 +3 → 11

Hint 2

The numbers 1, 4, 8, 11 satisfy the conditions of the problem.

Hint 3

You should repeat the pattern from Hint 1.

Hint 4

1, 4, 8, 11, 14, 18, 21, …, 2011, 2014, 2018, 2021, 2024

Hint 5

Ans ≥ 2020 10 × 3 + 2 = 608

Hint 6

We should prove that Ans ≤ 608

Hint 7

prove that from 1 to 10, you can select at most 3 numbers.

Hint 8

To prove Hint 7, note that from each group of 3 consecutive elements, we can select at most one number.

\underbrace{1,\,2,\,3}_{} \qquad \underbrace{4,\,5,\,6}_{} \qquad \underbrace{7,\,8,\,9}_{},\,10

Hint 9

If we select 4 numbers from 1 to 10 using Hint 8, we must select 10, and similarly, we must select 1. Then, by using the conditions of the problem:

1	2	3	4	5	6	7	8	9	10
✓	✗	✗		✗		✗	✗	✗	✓

Since we want to select 4 elements, we would also have to select 4 and 6, which is a contradiction. ✘

Hint 10

We can select at most two elements from {2021, 2022, 2023, 2024}

Hint 11

Ans ≤ 2020 10 × 3 + 2 = 608

Final Answer

Problem 21

The numbers, in order, of each row and the numbers, in order, of each column of a $5\times 5$ array of integers form an arithmetic progression of length 5. The numbers in positions $(5,5),\ (2,4),\ (4,3),\ (3,1)$ are $0,\,48,\,16,\,12$ , respectively. What number is in position $(1,2)$ ?

\begin{bmatrix} \cdot & ? & \cdot & \cdot & \cdot\\ \cdot & \cdot & \cdot & 48 & \cdot\\ 12 & \cdot & \cdot & \cdot & \cdot\\ \cdot & \cdot & 16 & \cdot & \cdot\\ \cdot & \cdot & \cdot & \cdot & 0 \end{bmatrix}

Hints & Final Answer

Hint 1

Let the common difference in the first row be

d

. Then the third row is an arithmetic progression:

\begin{bmatrix} \cdot & ? & \cdot & \cdot & \cdot\\ \cdot & \cdot & \cdot & 48 & \cdot\\ 12 & 12+d & 12+2d & 12+3d & 12+4d\\ \cdot & \cdot & 16 & \cdot & \cdot\\ \cdot & \cdot & \cdot & \cdot & 0 \end{bmatrix}

Hint 2

So the common difference in the third column is

16-(12+2d)=4-2d.

\begin{bmatrix} \cdot & ? & \cdot & \cdot & \cdot\\ \cdot & \cdot & \cdot & 48 & \cdot\\ 12 & 12+d & 12+2d & 12+3d & 12+4d\\ \cdot & \cdot & 16 & \cdot & \cdot\\ \cdot & \cdot & \cdot & \cdot & 0 \end{bmatrix}

Hint 3

Fill in the third column:

\begin{bmatrix} \cdot & ? & 4+6d & \cdot & \cdot\\ \cdot & \cdot & 8+4d & 48 & \cdot\\ 12 & 12+d & 12+2d & 12+3d & 12+4d\\ \cdot & \cdot & 16 & \cdot & \cdot\\ \cdot & \cdot & 20-2d & \cdot & 0 \end{bmatrix}

Hint 4

The common difference of the fourth column is

(12+3d)-48=3d-36.

\begin{bmatrix} \cdot & ? & 4+6d & \cdot & \cdot\\ \cdot & \cdot & 8+4d & 48 & \cdot\\ 12 & 12+d & 12+2d & 12+3d & 12+4d\\ \cdot & \cdot & 16 & \cdot & \cdot\\ \cdot & \cdot & 20-2d & \cdot & 0 \end{bmatrix}

Hint 5

Fill in the fourth column:

\begin{bmatrix} \cdot & ? & 4+6d & 84-3d & \cdot\\ \cdot & \cdot & 8+4d & 48 & \cdot\\ 12 & 12+d & 12+2d & 12+3d & 12+4d\\ \cdot & \cdot & 16 & -24+6d & \cdot\\ \cdot & \cdot & 20-2d & -60+9d & 0 \end{bmatrix}

Hint 6

The common difference of the fourth row is

(-24+6d)-16=-40+6d.

\begin{bmatrix} \cdot & ? & 4+6d & 84-3d & \cdot\\ \cdot & \cdot & 8+4d & 48 & \cdot\\ 12 & 12+d & 12+2d & 12+3d & 12+4d\\ \cdot & \cdot & 16 & -24+6d & \cdot\\ \cdot & \cdot & 20-2d & -60+9d & 0 \end{bmatrix}

Hint 7

Calculate the number in the fourth row and fifth column:

(4,5)=-64+12d.

\begin{bmatrix} \cdot & ? & 4+6d & 84-3d & \cdot\\ \cdot & \cdot & 8+4d & 48 & \cdot\\ 12 & 12+d & 12+2d & 12+3d & 12+4d\\ \cdot & \cdot & 16 & -24+6d & -64+12d\\ \cdot & \cdot & 20-2d & -60+9d & 0 \end{bmatrix}

Hint 8

By considering the fifth column:

\begin{aligned} -64+12d-(12+4d)&=0-(-64+12d)\\ 20d&=140\\ d&=7 \end{aligned}

\begin{bmatrix} \cdot & ? & 4+6d & 84-3d & \cdot\\ \cdot & \cdot & 8+4d & 48 & \cdot\\ 12 & 12+d & 12+2d & 12+3d & 12+4d\\ \cdot & \cdot & 16 & -24+6d & -64+12d\\ \cdot & \cdot & 20-2d & -60+9d & 0 \end{bmatrix}

Hint 9

By using

d=7

in the first row:

\begin{aligned} &?,\ 4+6(7),\ 84-3(7)\\ &?,\ 46,\ 63\\ 63-46&=46-?\\ ?&=29 \end{aligned}

Final Answer

Problem 22

Let $\mathcal{K}$ be the kite formed by joining two right triangles with legs $1$ and $\sqrt{3}$ along a common hypotenuse. Eight copies of $\mathcal{K}$ are used to form the polygon shown below. What is the area of triangle $\triangle ABC$ ?

$2 + 3\sqrt{3}$
$\frac{9}{2}\sqrt{3}$
$\frac{10 + 8\sqrt{3}}{3}$
8
$5\sqrt{3}$

Hints & Final Answer

Hint 1

Observe the construction of

\triangle ABC

using eight copies of

\mathcal{K}

Hint 2

The length of

\overline{AB}

is computed by summing the smaller segments along its base:

AB = \tfrac{1}{2} + 1 + 1 + \tfrac{1}{2} + \tfrac{3}{2} + \tfrac{3}{2} = 6.

Hint 3

The height perpendicular to

AB

h = \tfrac{3\sqrt{3}}{2}.

Final Answer

(B)

\frac{1}{2} \times AB \times h = \frac{1}{2} \times 6 \times \frac{3\sqrt{3}}{2} = \frac{9\sqrt{3}}{2}

Problem 23

Integers a, b, and c satisfy $ab+c=100$ , $bc+a=87$ , and $ca+b=60$ . What is $ab+bc+ca$ ?

Hints & Final Answer

Hint 1

Use problem 22. Approaches to similar equations.

Hint 2

ab + c = 100 \\ bc + a = 87 \\ \Rightarrow ab + c - bc - a = 13

Hint 3

b(a - c) + c - a = (a - c)(b - 1) = 13

Hint 4

b - 1 = \pm 1, \pm 13 \Rightarrow b = 0, 2, -12, 14

Hint 5

Case 1:

b = 0

ab + c = 100 \Rightarrow c = 100 \\ bc + a = 87 \Rightarrow a = 87 \\ ca + b = 60 \Rightarrow ca = 60

Contradiction.

Hint 6

Case 2:

b = 2

2a + c = 100 \Rightarrow 4a + 2c = 200 \\ 2c + a = 87 \Rightarrow 3a = 13

No integer solution.

Hint 7

Case 3:

b = 14

14a + c = 100 \Rightarrow 196a + 14c = 1400 \\ 14c + a = 87 \Rightarrow 195a = 1313

Not an integer.

Hint 8

Case 4:

b = -12

-12a + c = 100 \Rightarrow -144a + 12c = 1200 \\ -12c + a = 87 \Rightarrow -143a = 1287 \Rightarrow a = -9

Then

b = -12, c = -8

Final Answer

ab + bc + ca = (-9)(-12) + (-12)(-8) + (-8)(-9) = 276

(D) 276

Problem 24

A bee is moving in three-dimensional space. A fair six-sided die with faces labeled $A^+, A^-, B^+, B^-, C^+, C^-$ is rolled. Suppose the bee occupies the point $(a,b,c)$ . If the die shows $A^+$ , then the bee moves to the point $(a+1,b,c)$ , and if the die shows $A^-$ , then the bee moves to the point $(a-1,b,c)$ . Analogous moves are made with the other four outcomes. Suppose the bee starts at the point $(0,0,0)$ and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?

$\tfrac{1}{54}$
$\tfrac{7}{54}$
$\tfrac{1}{6}$
$\tfrac{5}{18}$
$\tfrac{2}{5}$

Hints & Final Answer

Hint 1

Calculate the total number of cases first.

Hint 2

N(\text{total}) = 6^4.

Hint 3

Calculate the number of favorable cases.

Hint 4

The first movement has 6 possible cases, and all of them are similar.

Hint 5

The second movement has 4 possible cases, and all of them are similar.

Hint 6

N(\text{favorable}) = 6 \times 4 \times ?

Hint 7

WLOG, consider:

(0,0,0)\to(1,0,0)\to(1,0,1).

Hint 8

For the third movement, there are three possible cases to go:

(1,1,1),\ (1,-1,1),\ (0,0,1).

The first two cases are similar.

Hint 9

Case 1: After the third movement, the position is either

(1,1,1)

(1,-1,1)

. WLOG, suppose it lands on

(1,1,1)

. Then there are two possible moves: to go left or down.

\Rightarrow\ 2 \times 2 = 4.

Hint 10

N(\text{favorable}) = 6 \times 4 \left(2 \times 2 + 1 \times 3\right) = 6 \times 4 \times 7.

Hint 11

Probability:

\dfrac{6 \times 4 \times 7}{6 \times 6 \times 6 \times 6} = \dfrac{7}{3 \times 3 \times 6} = \dfrac{7}{54}.

Final Answer

(B)

\dfrac{7}{54}

Problem 25

The figure below shows a dotted grid 8 cells wide and 3 cells tall consisting of $1'' \times 1''$ squares. Carl places 1-inch toothpicks along some of the sides of the squares to create a closed loop that does not intersect itself. The numbers in the cells indicate the number of sides of that square that are to be covered by toothpicks, and any number of toothpicks are allowed if no number is written. In how many ways can Carl place the toothpicks?