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

Latest revision as of 00:17, 12 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 $252/12 = \boxed{21}$ , 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 $a=\boxed{21}.$ ~samrocksnature

@@ Line 2: / Line 2: @@
 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>?
-==Solution==
+==Solution 1==
-asdf
+<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>, which works.
+~Bradygho
+==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>.
+(With <math>a=21</math>, we have <math>b=42, c=63, d=84,</math> which is valid)
+~Apple321
+==Solution 3 (A Little Bashy)==
+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>
+<math>\linebreak</math>
+~Geometry285
+==Solution 4 (Very algebraic)==
+If <math>a</math> divides <math>252,</math> then <math>3a</math> and <math>4a</math> must also divide <math>252.</math> This implies that <math>\frac{252}{3a},\frac{252}{4a}</math> are both integers, and that <math>a</math> divides and multiplying, we have that <math>a</math> divides <math>84</math> and <math>63.</math> The greatest common divisor of <math>84</math> and <math>63</math> is <math>21,</math> and we can check that indeed <math>a=\boxed{21}.</math> ~samrocksnature
+==See also==
+#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
+#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
+#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
+{{JMPSC Notice}}

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