Difference between revisions of "2021 JMC 10 Problems/Problem 11"
Skyscraper (talk | contribs) (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{(...") |
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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 < | + | 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 15:14, 1 April 2021
Problem
There exist positive integers 
that satisfy 
What is the sum of all possible values of 
Solution
Note that must be of the form 
where 
and 
To find this, observe that 
must divide 
Suppose that 
This implies that 
so 
must be a divisor of 
confirming what we noted. The sum of all equals ![\[3(1+2+4)(1+5) = 126.\]](http://latex.artofproblemsolving.com/f/4/9/f492409b79b6a5ae8d6c5d66073cb12679655087.png)
