Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 3"
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~Grisham | ~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== | ==See also== | ||
Latest revision as of 20:06, 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. 
