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Difference between revisions of "2022 AMC 10A Problems/Problem 7"

(→‎Solution 2: Fixed some errors and reformatted.)
(Merged solutions so that they are more self-explanatory. Retained credits of authors.)
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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>
  
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== Solution 1 ==  
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== Solution ==  
 
Note that
 
Note that
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 15: Line 15:
 
15 &= 3\cdot5.
 
15 &= 3\cdot5.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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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>
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Let <math>n = 2^a\cdot3^b\cdot5^c.</math> It follows that:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>From the least common multiple condition, we have <cmath>\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,</cmath> from which <math>a=2, b\in\{0,1,2\},</math> and <math>c=1.</math></li><p>
 +
  <li>From the greatest common divisor condition, we have <cmath>\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,</cmath> from which <math>b=1.</math></li><p>
 +
</ol>
 +
Together, we have <math>n=2^2\cdot3^1\cdot5^1=60.</math> The sum of its digits is <math>6+0=\boxed{\textbf{(B) } 6}.</math>
  
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From the greatest common divisor condition, we conclude that <math>k=1.</math>
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~MRENTHUSIASM ~USAMO333
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Therefore, we have <math>n=2^2\cdot 3^1\cdot5=60.</math> The sum of its digits is <math>6+0=\boxed{\textbf{(B) } 6}.</math>
 
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~MRENTHUSIASM
 
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== Solution 2 ==
 
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Since the LCM contains only factors of <math>2</math>, <math>3</math>, and <math>5</math>, <math>n</math> cannot be divisible by any other prime.
 
−
 
 
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Let <math>n = 2^a 3^b 5^c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are nonnegative integers.
 
−
 
 
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We know that <cmath>\operatorname{lcm}(n,18) = \operatorname{lcm}(n,2\cdot3^2) = \operatorname{lcm}(2^a \cdot 3^b \cdot 5^c, 2\cdot3^2) = 2^ {\max(a,1)} \cdot 3^ {\max(b,2)} \cdot 5^c = 180 = 2^2 \cdot 3^2 \cdot 5.</cmath>
 
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Thus,
 
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(1) <math>\max(a,1) = 2</math>, so <math>a = 2</math>.
 
−
 
 
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(2) <math>\max(b,2) = 2</math>, so <math>0 \le b \le 2</math>.
 
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(3) <math>c = 1</math>.
 
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From the GCD information, <cmath>\gcd(n,45) = \gcd(n, 3^2 \cdot 5) = \gcd(2^a \cdot 3^b \cdot 5^c, 3^2 \cdot 5)
 
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= 3^{\min(b,2)} \cdot 5^{\min(c,1)} = 15 = 3\cdot5.</cmath>
 
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This means, that since <math>c = 1</math>, it follows that <math>3^{\min(b,2)} \cdot 5 = 3\cdot5</math>, so <math>\min(b,2) = 1</math> and <math>b = 1</math>.
 
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Hence, multiplying using <math>a = 2</math>, <math>b = 1</math>, <math>c = 1</math> gives <math>n = 60</math> and the sum of digits is <math>\boxed{\textbf{(B) } 6}</math>.
 
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~USAMO333
 
  
 
==Video Solution 1 (Quick and Easy)==
 
==Video Solution 1 (Quick and Easy)==

Revision as of 23:12, 10 January 2023

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

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:

  1. 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 $c=1.$

  2. 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 have $n=2^2\cdot3^1\cdot5^1=60.$ The sum of its digits is $6+0=\boxed{\textbf{(B) } 6}.$

~MRENTHUSIASM ~USAMO333

Video Solution 1 (Quick and Easy)

https://youtu.be/YI1E8C3ZX-U

~Education, the Study of Everything

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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