1985 AHSME Problems/Problem 27
Problem
Consider a sequence defined by
and in general
for .
What is the smallest value of for which is an integer?
Solution
First, we will use induction to prove that
We see that . This is our base case.
Now, we have . Thus the induction is complete.
We now get rid of the cubed roots by introducing fractions into the exponents.
.
Notice that since isn't a perfect power, is integral if and only if the exponent, , is integral. By the same logic, this is integeral if and only if is integral. We can now clearly see that the smallest positive value of for which this is integral is .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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.