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

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==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 | ||

+ | |||

+ | ==Solution 2== | ||

+ | Suppose <math>x</math> is odd. We have <math>xk</math> for <math>k \equiv 0 \mod 2</math> must work for <math>xk \le 100</math>. Clearly <math>k=\{2,4,6,8,10 \}</math>, which means the maximum value that <math>xk</math> can take on is <math>90=9 \cdot 10</math>, and the minimum value it can take on is <math>2=2 \cdot 1</math>. Since we need <b>exactly 5</b> even integers, only <math>x=9</math> will work. Now, suppose <math>x</math> is even. We have <math>1 \le k \le 5</math>, which means <math>x=\{18,20 \}</math> hold exactly <math>5</math> even integer multiples. The answer is <math>18+20+9=\boxed{47}</math> | ||

+ | |||

+ | ~Geometry285 | ||

+ | |||

+ | ==See also== | ||

+ | #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]] | ||

+ | #[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]] | ||

+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||

+ | {{JMPSC Notice}} |

## Latest revision as of 21:06, 11 July 2021

## Contents

## 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

## Solution 2

Suppose is odd. We have for must work for . Clearly , which means the maximum value that can take on is , and the minimum value it can take on is . Since we need **exactly 5** even integers, only will work. Now, suppose is even. We have , which means hold exactly even integer multiples. The answer is

~Geometry285

## See also

- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.