Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 3"
(→Solution) |
|||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
There are exactly <math>5</math> even positive integers less than or equal to <math>100</math> that are divisible by <math>x</math>. What is the sum of all possible positive integer values of <math>x</math>? | There are exactly <math>5</math> even positive integers less than or equal to <math>100</math> that are divisible by <math>x</math>. What is the sum of all possible positive integer values of <math>x</math>? | ||
| + | |||
| + | ==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> | ||
Revision as of 14:33, 11 July 2021
Problem
There are exactly 
even positive integers less than or equal to 
that are divisible by 
. What is the sum of all possible positive integer values of ?
Solution
must have exactly 5 even multiples less than . We have two cases, either is odd or even. If is even, then 
. We solve the inequality to find 
, but since must be an integer we have x = 18, 20. If is odd, then we can set up the inequality 
. Solving for the integers must be 
. The sum is 
or 
