Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 3"

m (Solution)
Line 4: Line 4:
 
==Solution==
 
==Solution==
 
<math>x</math> must have exactly 5 even multiples less than <math>100</math>. We have two cases, either <math>x</math> is odd or even. If <math>x</math> is even, then <math>5x < 100 < 6x</math>. We solve the inequality to find <math>\frac{50}{3} \leq x \leq 20</math>, but since <math>x</math> must be an integer we have x = 18, 20. If <math>x</math> is odd, then we can set up the inequality <math>10x\leq100\leq12x</math>. Solving for the integers <math>x</math> must be <math>9</math>. The sum is <math>18+20+9</math> or <math>\boxed{47}</math>
 
<math>x</math> must have exactly 5 even multiples less than <math>100</math>. We have two cases, either <math>x</math> is odd or even. If <math>x</math> is even, then <math>5x < 100 < 6x</math>. We solve the inequality to find <math>\frac{50}{3} \leq x \leq 20</math>, but since <math>x</math> must be an integer we have x = 18, 20. If <math>x</math> is odd, then we can set up the inequality <math>10x\leq100\leq12x</math>. Solving for the integers <math>x</math> must be <math>9</math>. The sum is <math>18+20+9</math> or <math>\boxed{47}</math>
 +
 +
~Grisham

Revision as of 15:35, 11 July 2021

Problem

There are exactly $5$ even positive integers less than or equal to $100$ that are divisible by $x$. What is the sum of all possible positive integer values of $x$?

Solution

$x$ must have exactly 5 even multiples less than $100$. We have two cases, either $x$ is odd or even. If $x$ is even, then $5x < 100 < 6x$. We solve the inequality to find $\frac{50}{3} \leq x \leq 20$, but since $x$ must be an integer we have x = 18, 20. If $x$ is odd, then we can set up the inequality $10x\leq100\leq12x$. Solving for the integers $x$ must be $9$. The sum is $18+20+9$ or $\boxed{47}$

~Grisham

Invalid username
Login to AoPS