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

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(→‎Solution 2: Fixed some errors and reformatted.)
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== Solution 2 ==
 
== Solution 2 ==
−
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.
+
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.
−
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>
+
Let <math>n = 2^a 3^b 5^c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are nonnegative integers.
  
−
(2) <math>max(a,2)</math> = <math>2</math> so <math>0 \le a \le 2</math>  
+
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>
 +
Thus,
  
−
(3) <math>c = 1.</math>
+
(1) <math>\max(a,1) = 2</math>, so <math>a = 2</math>.
  
−
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>
+
(2) <math>\max(b,2) = 2</math>, so <math>0 \le b \le 2</math>.
−
= <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>.  
+
(3) <math>c = 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>.
+
 
 +
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)  
 +
= 3^{\min(b,2)} \cdot 5^{\min(c,1)} = 15 = 3\cdot5.</cmath>
 +
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>.  
 +
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>.
  
 
~USAMO333
 
~USAMO333

Revision as of 21:09, 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 1

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 $k=1.$

Therefore, we have $n=2^2\cdot 3^1\cdot5=60.$ The sum of its digits is $6+0=\boxed{\textbf{(B) } 6}.$

~MRENTHUSIASM

Solution 2

Since the LCM contains only factors of $2$, $3$, and $5$, $n$ cannot be divisible by any other prime.

Let $n = 2^a 3^b 5^c$, where $a$, $b$, and $c$ are nonnegative integers.

We know that \[\operatorname{lcm}(n,18) = \operatorname{lcm}(n,2\cdot3^2) = \operatorname{lcm}(2^a \cdot 3^b \cdot 5^c, 2\cdot3^2) = 2^ {\max(a,1)} \cdot 3^ {\max(b,2)} \cdot 5^c = 180 = 2^2 \cdot 3^2 \cdot 5.\] Thus,

(1) $\max(a,1) = 2$, so $a = 2$.

(2) $\max(b,2) = 2$, so $0 \le b \le 2$.

(3) $c = 1$.

From the GCD information, \[\gcd(n,45) = \gcd(n, 3^2 \cdot 5) = \gcd(2^a \cdot 3^b \cdot 5^c, 3^2 \cdot 5) = 3^{\min(b,2)} \cdot 5^{\min(c,1)} = 15 = 3\cdot5.\] This means, that since $c = 1$, it follows that $3^{\min(b,2)} \cdot 5 = 3\cdot5$, so $\min(b,2) = 1$ and $b = 1$. Hence, multiplying using $a = 2$, $b = 1$, $c = 1$ gives $n = 60$ and the sum of digits is $\boxed{\textbf{(B) } 6}$.

~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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