1968 AHSME Problems/Problem 33

Problem

A number $N$ has three digits when expressed in base $7$ . When $N$ is expressed in base $9$ the digits are reversed. Then the middle digit is:

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$

Solution

Call the number abc. Then, 49a+7b+c=81c+9b+a. (Breaking down the number in base-form) Moving everything to one side, 48a=2b+80c, b=40c-24a=8(5c-2a). This shows that b is a multiple of 8. Thus, b=0 (since 8>7, the base of base 7). So b=0. Select "A." I have no idea how to format, so please help. Thanks.

$\fbox{A}$

1968 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 32	Followed by Problem 34
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1968 AHSME Problems/Problem 33

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