Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 11"
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==Solution 4 (Very algebraic)== | ==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> | 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> | ||
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+ | ==See also== | ||
+ | #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]] | ||
+ | #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Revision as of 16:19, 11 July 2021
Contents
[hide]Problem
If and , , , and are divisors of , what is the maximum value of ?
Solution 1
must be a number such that , , . Thus, we must have . This implies the maximum value of is , which works.
~Bradygho
Solution 2
Notice that . Because and it is invalid for to be a multiple of . With similar reasoning, must have at most one factor of . Thus, .
(With , we have which is valid)
~Apple321
Solution 3 (A Little Bashy)
Note , so the divisors are . We see the set is the largest 4-digit set we can form, so the answer is ~Geometry285
Solution 4 (Very algebraic)
If divides then and must also divide This implies that are both integers, and that divides and multiplying, we have that divides and The greatest common divisor of and is and we can check that indeed
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.