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 $\overline{abc}$ in base 7.

Then, $49a+7b+c=81c+9b+a$ . (Breaking down the number in base-form).

After combining like terms and moving the variables around, $48a=2b+80c$ , $b=40c-24a=8(5c-2a)$ . This shows that $b$ is a multiple of 8 (we only have to find the middle digit under one of the bases). Thus, $b=0$ (since 8>6, the largest digit in base 7).

Select $\fbox{A}$ as our answer.

~hastapasta

Solution 2

If the number in base 7 has reverse digits in base 9, then $49a+7b+c=81c+9b+a$ . When 3-digit numbers' digits reverse, the first and third digits swap places and the middle one doesn't change. So, $7b=9b$ and therefore, $b=0$ .

The correct choice is $\fbox{A}$ .

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

Contents

Problem

Solution

Solution 2

See also