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jmc

number theory intermediate

Problem

Suppose

65 6_{7} = 3 a b_{10}

, where

a

and

b

represent base-10 digits. Find

\frac{a \cdot b}{15}

.

Solution

Note that

65 6_{7} = 6 \cdot 7^{2} + 5 \cdot 7^{1} + 6 \cdot 7^{0} = 33 5_{10}

. Therefore,

a = 3

,

b = 5

, and

\frac{a \cdot b}{15} = \frac{3 \cdot 5}{15} = 1

.

Final answer

1