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Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 11"

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==Solution==
 
==Solution==
 
<math>a</math> must be a number such that <math>2a \mid 252</math>, <math>3a \mid 252</math>, <math>4a \mid 252</math>. Thus, we must have <math>12a \mid 252</math>. This implies the maximum value of <math>a</math> is <math>252/12 = \boxed{21}</math>
 
<math>a</math> must be a number such that <math>2a \mid 252</math>, <math>3a \mid 252</math>, <math>4a \mid 252</math>. Thus, we must have <math>12a \mid 252</math>. This implies the maximum value of <math>a</math> is <math>252/12 = \boxed{21}</math>
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~Bradygho

Revision as of 22:38, 10 July 2021

Problem

If $a : b : c : d=1 : 2 : 3 : 4$ and $a$, $b$, $c$, and $d$ are divisors of $252$, what is the maximum value of $a$?

Solution

$a$ must be a number such that $2a \mid 252$, $3a \mid 252$, $4a \mid 252$. Thus, we must have $12a \mid 252$. This implies the maximum value of $a$ is $252/12 = \boxed{21}$

~Bradygho

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