Difference between revisions of "2021 JMC 10 Problems/Problem 11"

(Created page with "==Problem== There exist positive integers <math>k</math> that satisfy <math>k = 3\gcd(20,k).</math> What is the sum of all possible values of <math>k?</math> <math>\textbf{(...")
 
(Solution)
 
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==Solution==
 
==Solution==
Note that <math>k</math> must be of the form <math>3 \cdot 2^a \cdot 5^b</math> where <math>a = 0,1,2</math> and <math>b = 0,1.</math> To find this, observe that <math>3</math> must divide <math>k.</math> Suppose that <math>k=3x.</math> This implies that <math>x = \gcd(k,20),</math> so <math>x</math> must be a divisor of <math>20,</math> confirming what we noted. The sum of all <math>k</math> equals <math>3(1+2+4)(1+5) = 126.</math>
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Note that <math>k</math> must be of the form <math>3 \cdot 2^a \cdot 5^b</math> where <math>a = 0,1,2</math> and <math>b = 0,1.</math> To find this, observe that <math>3</math> must divide <math>k.</math> Suppose that <math>k=3x.</math> This implies that <math>x = \gcd(k,20),</math> so <math>x</math> must be a divisor of <math>20,</math> confirming what we noted. The sum of all <math>k</math> equals <cmath>3(1+2+4)(1+5) = 126.</cmath>

Latest revision as of 16:14, 1 April 2021

Problem

There exist positive integers $k$ that satisfy $k = 3\gcd(20,k).$ What is the sum of all possible values of $k?$

$\textbf{(A) } 84 \qquad\textbf{(B) } 108 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 126 \qquad\textbf{(E) } 132$

Solution

Note that $k$ must be of the form $3 \cdot 2^a \cdot 5^b$ where $a = 0,1,2$ and $b = 0,1.$ To find this, observe that $3$ must divide $k.$ Suppose that $k=3x.$ This implies that $x = \gcd(k,20),$ so $x$ must be a divisor of $20,$ confirming what we noted. The sum of all $k$ equals \[3(1+2+4)(1+5) = 126.\]