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 cube 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 integral 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) | ||
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Followed by Problem 28 | |
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