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2022 AMC 10A Problems/Problem 7

Revision as of 00:46, 12 November 2022 by Usamo333 (talk | contribs) (→‎Solution 2)

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 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 $2^2\cdot 3^1\cdot5=60.$ 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 $lcm(n, 18)$ = $lcm(n, 3^2 * 2)$ = $lcm(2^a * 3^b * 5^c, 3^2 * 2)$ = $lcm(2^ {max(a,1)} \cdot 3^ {max(a,2)} \cdot 5^b)$ = $180$ = $2^2 \cdot 3^2 \cdot 5.$ Thus 

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

(2) $max(a,2)$ = $2$ so $0$ <= $a$ <= $2$ 

(3) $c = 1.$

From the gcf information, $gcf(n,45)$ = gcf(n, $3^2 \cdot 5$) = $gcf(2^a \cdot 3^b \cdot 5^c, 3^2 \cdot 5)$ = $3^{min(b,2)} \cdot 5^{min(c,1)} = 15$ This means, that since $c = 1$, $3^{min(b,2)} \cdot 5 = 15$, 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 hence $\boxed{\textbf{(B)} ~6}$.

~USAMO333

See Also

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

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