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 $\begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*}$ Let $n = 2^a\cdot3^b\cdot5^c.$ It follows that:

From the least common multiple condition, we have $\operatorname{lcm}(n,18) = \operatorname{lcm}(2^a\cdot3^b\cdot5^c,2\cdot3^2) = 2^{\max(a,1)}\cdot3^{\max(b,2)}\cdot5^{\max(c,0)} = 2^2\cdot3^2\cdot5,$ from which $a=2, b\in\{0,1,2\},$ and $c=1.$

From the greatest common divisor condition, we have $\gcd(n,45) = \gcd(2^2\cdot3^b\cdot5,3^2\cdot5) = 2^{\min(2,0)}\cdot3^{\min(b,2)}\cdot5^{\min(1,1)} = 3\cdot5,$ from which $b=1.$

Together, we conclude that $n=2^2\cdot3\cdot5=60.$ The sum of its digits is $6+0=\boxed{\textbf{(B) } 6}.$

~MRENTHUSIASM ~USAMO333

Solution 2

The options for $\text{lcm}(x, 18)=180$ are $20$ , $60$ , and $180$ . The options for $\text{gcd}(y, 45)=15$ are $15$ , $30$ , $60$ , $75$ , etc. We see that $60$ appears in both lists, therefore, $6+0=\boxed{\textbf{(B) } 6}$ .

~MrThinker

Remark

If you ignore or mess up the LCM, and get $n=15$ , you'll still get the correct answer.

Video Solution 1

https://youtu.be/YI1E8C3ZX-U

~Education, the Study of Everything

Video Solution 2

https://youtu.be/q2y-Wfdi4q8

~savannahsolver

Video Solution 3 (Smart and Simple)

https://youtu.be/7yAh4MtJ8a8?si=o-aImrVOwbH1HoZv&t=673

~Math-X

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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