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

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<math>(3) c = 1.</math>
 
<math>(3) c = 1.</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>
 
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>
−
= <math>3^min(b,2) \cdot 5^min(c,1) = 15</math>
+
= <math>3^{min(b,2)} \cdot 5^{min(c,1)} = 15</math>
−
This means, that since c = 1, 3^min(b,2) \cmath 5 = 15, so min(b,2) = 1 and b = 1.  
+
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 a = 2, b = 1, c = 1 gives n = 60 and the sum of digits is hence 6.
+
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 ==

Revision as of 01:41, 12 November 2022

Problem

The least common multiple of a positive divisor $n$ 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 that $k=1.$

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

~MRENTHUSIASM

Solution 2

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 $n$ ?


$\textbf{(A)} 3 \textbf{ (B)} 6 \textbf{ (C)} 8 \textbf{ (D)} 9 \textbf{ (E)} 12$

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 $lcm(n, 18)$ = $lcm(n, 3^2 * 2)$ = $lcm(2^a * 3^b * 5^c, 3^2 * 2)$ = $lcm(2^ {max(a,1)} \cdot 3^ {max(a,2)} \cdot 5^b)$ = $180$ = $2^2 \cdot 3^2 \cdot 5.$ Thus

$(1) max(a,1) = 2 -> a = 2$

$(2) max(a,2) = 2 -> 0 <= a <= 2$

$(3) c = 1.$ From the gcf information, $gcf(n,45)$ = gcf(n, $3^2 \cdot 5$) = $gcf(2^a \cdot 3^b \cdot 5^c, 3^2 \cdot 5)$ = $3^{min(b,2)} \cdot 5^{min(c,1)} = 15$ This means, that since $c = 1$, $3^{min(b,2)} \cdot 5 = 15$, 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 hence 6.

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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