2022 AMC 10A Problems/Problem 7

Problem

The least common multiple of a positive divisor $n$ 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

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 that $k=1.$

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

~MRENTHUSIASM

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

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

Page

Toolbox

Search

2022 AMC 10A Problems/Problem 7

Problem

Solution

See Also