Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 2)
 
(5 intermediate revisions by 4 users not shown)
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>
 +
 +
~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 20:06, 11 July 2021

Problem

There are exactly $5$ even positive integers less than or equal to $100$ that are divisible by $x$. What is the sum of all possible positive integer values of $x$?

Solution

$x$ must have exactly 5 even multiples less than $100$. We have two cases, either $x$ is odd or even. If $x$ is even, then $5x < 100 < 6x$. We solve the inequality to find $\frac{50}{3} \leq x \leq 20$, but since $x$ must be an integer we have x = 18, 20. If $x$ is odd, then we can set up the inequality $10x\leq100\leq12x$. Solving for the integers $x$ must be $9$. The sum is $18+20+9$ or $\boxed{47}$

~Grisham

Solution 2

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

~Geometry285

See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_3&oldid=158144"