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2022 AMC 10A Problems/Problem 7: Difference between revisions

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==Problem 7==
==Problem==


The least common multiple of a positive divisor <math>n</math> and <math>18</math> is <math>180</math>, and the greatest common divisor of <math>n</math> and <math>45</math> is <math>15</math>. What is the sum of the digits of <math>n</math>?
The least common multiple of a positive divisor <math>n</math> and <math>18</math> is <math>180</math>, and the greatest common divisor of <math>n</math> and <math>45</math> is <math>15</math>. What is the sum of the digits of <math>n</math>?
Line 5: Line 5:
<math>\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12</math>
<math>\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12</math>


[[2022 AMC 10A Problems/Problem 7|Solution]]
== Solution ==
Note that
<cmath>\begin{align*}
18 &= 2\cdot3^2, \\
180 &= 2^2\cdot3^2\cdot5, \\
45 &= 3^2\cdot5 \\
15 &= 3\cdot5.
\end{align*}</cmath>
From the least common multiple condition, we conclude that <math>n=2^2\cdot 3^k\cdot5,</math> where <math>k\in\{0,1,2\}.</math>
 
From the greatest common divisor condition, we conclude that that <math>k=1.</math>
 
Therefore, we have <math>2^2\cdot 3^1\cdot5=60.</math> The sum of its digits is <math>6+0=\boxed{\textbf{(B) } 6}.</math>
 
~MRENTHUSIASM
 
== See Also ==
 
{{AMC10 box|year=2022|ab=A|num-b=5|num-a=7}}
{{MAA Notice}}

Revision as of 21:41, 11 November 2022

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

See Also

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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