1966 AHSME Problems/Problem 39
Problem
In base 
the expanded fraction 
becomes 
, and the expanded fraction 
becomes 
. In base 
fraction , when expanded, becomes 
, while the fraction becomes 
. The sum of and , each written in the base ten, is:
Solution
First, let's write as a proper fraction in base . To do that, note that:
Multiplying this equation on both sides 
, we get:
Subtracting the first equation from the second one, we get:
Using a very similar method as above, we can see that:
and 
.
Dividing the 2 equations to get some potential solutions, we get (note that we don't have to worry about division by zero because 
):
Since non-integer bases are rarely used, we can try to assume that the solution is positive integers and see if we get any solutions.
Notice that 
has to be a negative factor of 840. We need to plug in values of 
. 840 divides -21, so we plug in 8 to check. Luckily, when 
, we see that 
, and furthermore, 11+8=19 is one of the answers. We can quickly test it into the original equations and , and we see they indeed work. Therefore the answer is 
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 40 | |
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