Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 7"
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+ | Clearly we see <math>C=1</math> does not work, but <math>C=5</math> works with simple guess-and-check. We have <math>AB5=\frac{795}{3}=265</math>, so <math>A=2</math> and <math>B=6</math>. The answer is <math>3(2)+6(2)+1(5)=\boxed{23}</math> | ||
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+ | ~Geometry285 |
Revision as of 11:25, 11 July 2021
Problem
If , , and each represent a single digit and they satisfy the equation find .
Solution
Notice that can only be and . However, is not divisible by , so the number . Thus,
~Bradygho
Solution 2
Clearly we see does not work, but works with simple guess-and-check. We have , so and . The answer is
~Geometry285