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

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:

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 $c=1.$

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

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

@@ Line 15: / Line 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>
+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>
-From the greatest common divisor condition, we conclude that <math>k=1.</math>
+~MRENTHUSIASM ~USAMO333
-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>
+==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>.
-~MRENTHUSIASM
+~MrThinker
-== Solution 2 ==
+== Remark ==
-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
-  (1) <math> max(a,1)</math> = <math>2</math> so <math>a = 2 </math>
+If you ignore or mess up the LCM, and get  <math>n=15</math>, you'll still get the correct answer.
- (2) <math>max(a,2)</math> = <math>2</math> so <math>0 \le a \le 2</math>
+==Video Solution 1 ==
+https://youtu.be/YI1E8C3ZX-U
- (3) <math>c = 1.</math>
+~Education, the Study of Everything
-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>
+==Video Solution 2==
-= <math>3^{min(b,2)} \cdot 5^{min(c,1)} = 15</math>
+https://youtu.be/q2y-Wfdi4q8
-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 <math>n = 60</math> and the sum of digits is hence <math>\boxed{\textbf{(B)} ~6}</math>.
-~USAMO333
+~savannahsolver
-==Video Solution 1 (Quick and Easy)==
+==Video Solution 3 (Smart and Simple) ==
-https://youtu.be/YI1E8C3ZX-U
+https://youtu.be/7yAh4MtJ8a8?si=o-aImrVOwbH1HoZv&t=673
+~Math-X
-~Education, the Study of Everything
+==Video Solution 4==
+https://youtu.be/5KAiNlqbrsQ
 == See Also ==

Page

Toolbox

Search