1989 AIME Problems/Problem 4
Problem
If 
are consecutive positive integers such that 
is a perfect square and 
is a perfect cube, what is the smallest possible value of 
?
Solution
Since the middle term of an arithmetic progression with an odd number of terms is the average of the series, we know 
and 
. Thus, must be in the form of 
based upon the first part and in the form of 
based upon the second part, with 
and 
denoting an integers. is minimized if it’s prime factorization contains only 
, and since there is a cubed term in , 
must be a factor of . 
, which works as the solution.
Solution 2
Let 
, , 
, and 
equal 
, 
, 
, and 
, respectively. Call the square and cube 
and 
, where both k and m are integers. Then:
Now we know is a multiple of 125 and 
is a multiple of 5. The lower is, the lower the value of will be. Start from 5 and add 5 each time.
gives no solution for k
gives no solution for k
gives a solution for k.
-jackshi2006
Solution 3
Let the numbers be 
When then know 
is a perfect cube and 
is perfect cube. Since is divisible by 
we know that 
since otherwise we get a contradiction. This means 
in which plugging into the other expression we know 
is a perfect square. We know 
so we let 
to obtain the perfect square. This means that 
See also
| 1989 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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