1968 AHSME Problems/Problem 33
Problem
A number 
has three digits when expressed in base 
. When is expressed in base 
the digits are reversed. Then the middle digit is:
Solution
Call the number abc.
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).
So b=0.
Select 
as our answer.
I have no idea how to format besides the basics, so please help. Feel free to message me tips on AOPS. Thanks.
~hastapasta
See also
| 1968 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 34 | |
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