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Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 3"
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~Grisham
 
~Grisham
  
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==Solution 2==
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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>
  
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~Geometry285
  
 
==See also==
 
==See also==

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 [b]exactly 5[/b] 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

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