I'm thinking of a number, x, whose last 3 digits are somewhere between 400 and 500.
You could probably guess from those digits that it might be a square or a cube, but in fact it turns out to be both, and in fact the smallest possible such number.
What are the prime factors of x?
Extra credit: feeling adventurous, I next computed the smallest number, y, having the same last 3 digits, that was a seventh power, as well as a square and a cube.
What are the prime factors of y?
To find x it suffices to find the first 6th power whose hundreds place digit is 4.
A quick search of 6th powers finds 9^6 = 531441 = 729^2 = 81^3
The only prime factor of x is 3. Its prime factorization is 3^12.
To find y we need to find a 42nd power that ends in 441.
I was hoping to find patterns in 2nd, 3rd, 6th, 7th powers.
There are two patterns for 2nd powers:
(250n+21)^2 ends in 441 and (250n+229)^2 ends in 441
But I found no cubes besides 81^3, no 6th powers beyond 9^3, no 7th powers at all.
I may be on the wrong track. It seems if a number exists there must be a way to construct it. A search isn't working.
Posted by Jer
on 2013-01-30 14:33:57