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 14:35, 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