2021 JMPSC Accuracy Problems/Problem 11

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 $252/12 = \boxed{21}$ , which works.

~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

Solution 4 (Very algebraic)

If $a$ divides $252,$ then $3a$ and $4a$ must also divide $252.$ This implies that $\frac{252}{3a},\frac{252}{4a}$ are both integers, and that $a$ divides and multiplying, we have that $a$ divides $84$ and $63.$ The greatest common divisor of $84$ and $63$ is $21,$ and we can check that indeed $a=\boxed{21}.$

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