Difference between revisions of "2022 AMC 10A Problems/Problem 7"
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| − | ==Problem | + | ==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>? | ||
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<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> | ||
| − | + | == 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 
and 
is 
, and the greatest common divisor of and 
is 
. What is the sum of the digits of ?
Solution
Note that
From the least common multiple condition, we conclude that 
where 
From the greatest common divisor condition, we conclude that that 
Therefore, we have 
The sum of its digits is 
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
