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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.