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

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If <math>a : b : c : d=1 : 2 : 3 : 4</math> and <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are divisors of <math>252</math>, what is the maximum value of <math>a</math>?
 
If <math>a : b : c : d=1 : 2 : 3 : 4</math> and <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are divisors of <math>252</math>, what is the maximum value of <math>a</math>?
  
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==Solution #1==
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==Solution 1==
 
<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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==Solution #2==
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==Solution 2==
 
Notice that <math>252=2^2\cdot 3^2\cdot 7</math>. Because <math>b=2a</math> and <math>d=4a,</math> it is invalid for <math>a</math> to be a multiple of <math>2</math>. With similar reasoning, <math>a</math> must have at most one factor of <math>3</math>. Thus, <math>a=\boxed{21}</math>.
 
Notice that <math>252=2^2\cdot 3^2\cdot 7</math>. Because <math>b=2a</math> and <math>d=4a,</math> it is invalid for <math>a</math> to be a multiple of <math>2</math>. With similar reasoning, <math>a</math> must have at most one factor of <math>3</math>. Thus, <math>a=\boxed{21}</math>.
  
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~Apple321
 
~Apple321
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==Solution 3 (A Little Bashy)==
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Note <math>252=2^2 \cdot 3^2 \cdot 7</math>, so the divisors are <math>\{1,2,3,4,6,7,9,12,14,18,21,28,36,42,63,84,126,252 \}</math>. We see the set <math>\{21,42,63,84 \}</math> is the largest 4-digit set we can form, so the answer is <math>a=\boxed{21}</math>
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<math>\linebreak</math>
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~Geometry285

Revision as of 12:30, 11 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 1

$a$ 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


Solution 2

Notice that $252=2^2\cdot 3^2\cdot 7$. Because $b=2a$ and $d=4a,$ it is invalid for $a$ to be a multiple of $2$. With similar reasoning, $a$ must have at most one factor of $3$. Thus, $a=\boxed{21}$.


(With $a=21$, we have $b=42, c=63, d=84,$ which is valid)

~Apple321

Solution 3 (A Little Bashy)

Note $252=2^2 \cdot 3^2 \cdot 7$, so the divisors are $\{1,2,3,4,6,7,9,12,14,18,21,28,36,42,63,84,126,252 \}$. We see the set $\{21,42,63,84 \}$ is the largest 4-digit set we can form, so the answer is $a=\boxed{21}$ $\linebreak$ ~Geometry285

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