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

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<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> | ||

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+ | ~Grisham |

## Revision as of 15:35, 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

~Grisham