2008 Mock ARML 1 Problems/Problem 7

Problem

Compute the number of $3$-digit base-$5$ positive integer multiples of $7$ that are also divisible by $7$ when read in base $10$ instead of base $5$.

Solution

Let the number be $\overline{abc}$. Then $7 | 25a + 5b + c, 100a + 10b + c$, and so it must divide their difference, so $7 | 75a + 5b \Longrightarrow 7|-2(a+b)$, from which it follows that $7|a+b$. However, as $a,b < 5$, we have $\{a,b\} = \{3,4\}$, leading to $\boxed{2}$ solutions: $343, 434$.

See also

2008 Mock ARML 1 (Problems, Source)
Preceded by
Problem 6
Followed by
Problem 8
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