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

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{{duplicate|[[2022 AMC 10A Problems/Problem 7|2022 AMC 10A #7]] and [[2022 AMC 12A Problems/Problem 4|2022 AMC 12A #4]]}}
==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 integer <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>?


<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>
Line 13: Line 15:
15 &= 3\cdot5.
15 &= 3\cdot5.
\end{align*}</cmath>
\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>
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 conclude that <math>n=2^2\cdot3\cdot5=60.</math> The sum of its digits is <math>6+0=\boxed{\textbf{(B) } 6}.</math>
 
~MRENTHUSIASM ~USAMO333


From the greatest common divisor condition, we conclude that that <math>k=1.</math>
==Solution 2==
The options for <math>\text{lcm}(x, 18)=180</math> are <math>20</math>, <math>60</math>, and <math>180</math>. The options for <math>\text{gcd}(y, 45)=15</math> are <math>15</math>, <math>30</math>, <math>60</math>, <math>75</math>, etc. We see that <math>60</math> appears in both lists; therefore, <math>6+0=\boxed{\textbf{(B) } 6}</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>
~MrThinker


~MRENTHUSIASM
== Remark ==


== Solution 2 ==
If you ignore or mess up the LCM, and get  <math>n=15</math>, you'll still get the correct answer.
==Problem==


The least common multiple of a positive integer <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> ?
==Video Solution 1 ==
https://youtu.be/YI1E8C3ZX-U


~Education, the Study of Everything


<math>\textbf{(A)} 3  \textbf{ (B)} 6  \textbf{ (C)} 8  \textbf{ (D)} 9  \textbf{ (E)} 12</math>
==Video Solution 2==
https://youtu.be/q2y-Wfdi4q8


== Solution 2 ==
~savannahsolver
Since the <math>lcm</math> contains only factors of <math>2</math>, <math>3</math>, and <math>5</math>, <math>n</math> cannot be divisible by any other prime.
Let n = <math>2^a</math> <math>3^b</math> <math>5^c</math>, where <math> a</math> ,<math>b</math>, and <math>c</math> are nonnegative integers.
We know that <math>lcm(n, 18)</math> = <math>lcm(n, 3^2 * 2)</math> = <math>lcm(2^a * 3^b * 5^c, 3^2 * 2)</math> =
<math>lcm(2^ {max(a,1)} \cdot 3^ {max(a,2)} \cdot 5^b)</math> = <math>180</math> = <math>2^2 \cdot 3^2 \cdot 5.</math>
Thus


<math> (1) max(a,1) = 2 -> a = 2 </math>
==Video Solution 3 (Smart and Simple) ==
https://youtu.be/7yAh4MtJ8a8?si=o-aImrVOwbH1HoZv&t=673


<math> (2) max(a,2) = 2 -> a ∈ {0,1,2} </math>
~Math-X


<math>(3) c = 1.</math>
==Video Solution 4==
From the gcf information, <math>gcf(n,45)</math> = gcf(n, <math>3^2 \cdot 5</math>) = <math>gcf(2^a \cdot 3^b \cdot 5^c, 3^2 \cdot 5) </math>
https://youtu.be/5KAiNlqbrsQ
= <math>3^{min(b,2)} \cdot 5^{min(c,1)} = 15</math>
This means, that since <math>c = 1</math>, <math> 3^{min(b,2)} \cdot 5 = 15</math>, so <math>min(b,2)</math> = <math>1</math> and <math>b = 1</math>.  
Hence, multiplying using <math>a = 2</math>, <math>b = 1</math>, <math>c = 1</math> gives n = 60 and the sum of digits is hence 6.


== See Also ==
== See Also ==
{{AMC10 box|year=2022|ab=A|num-b=6|num-a=8}}
{{AMC10 box|year=2022|ab=A|num-b=6|num-a=8}}
{{AMC12 box|year=2022|ab=A|num-b=3|num-a=5}}
{{AMC12 box|year=2022|ab=A|num-b=3|num-a=5}}
[[Category:Introductory Number Theory Problems]]
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 12:29, 8 February 2024

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:

  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 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

Video Solution 4

https://youtu.be/5KAiNlqbrsQ

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

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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