2022 AMC 10A Problems/Problem 7

The following problem is from both the 2022 AMC 10A #7 and 2022 AMC 12A #4, so both problems redirect to this page.

Problem

The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$ ?

$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$

Solution 1

Note that $\begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*}$ From the least common multiple condition, we conclude that $n=2^2\cdot 3^k\cdot5,$ where $k\in\{0,1,2\}.$

From the greatest common divisor condition, we conclude that $k=1.$

Therefore, we have $n=2^2\cdot 3^1\cdot5=60.$ The sum of its digits is $6+0=\boxed{\textbf{(B) } 6}.$

~MRENTHUSIASM

Solution 2

Since the LCM contains only factors of $2$ , $3$ , and $5$ , $n$ cannot be divisible by any other prime.

Let $n = 2^a 3^b 5^c$ , where $a$ , $b$ , and $c$ are nonnegative integers.

We know that $\operatorname{lcm}(n,18) = \operatorname{lcm}(n,2\cdot3^2) = \operatorname{lcm}(2^a \cdot 3^b \cdot 5^c, 2\cdot3^2) = 2^ {\max(a,1)} \cdot 3^ {\max(b,2)} \cdot 5^c = 180 = 2^2 \cdot 3^2 \cdot 5.$ Thus,

(1) $\max(a,1) = 2$ , so $a = 2$ .

(2) $\max(b,2) = 2$ , so $0 \le b \le 2$ .

(3) $c = 1$ .

From the GCD information, $\gcd(n,45) = \gcd(n, 3^2 \cdot 5) = \gcd(2^a \cdot 3^b \cdot 5^c, 3^2 \cdot 5) = 3^{\min(b,2)} \cdot 5^{\min(c,1)} = 15 = 3\cdot5.$ This means, that since $c = 1$ , it follows that $3^{\min(b,2)} \cdot 5 = 3\cdot5$ , so $\min(b,2) = 1$ and $b = 1$ . Hence, multiplying using $a = 2$ , $b = 1$ , $c = 1$ gives $n = 60$ and the sum of digits is $\boxed{\textbf{(B) } 6}$ .

~USAMO333

Video Solution 1 (Quick and Easy)

https://youtu.be/YI1E8C3ZX-U

~Education, the Study of Everything

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2022 AMC 10A Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

Video Solution 1 (Quick and Easy)

See Also