You are not logged in.
Pages: 1
How many ways are there to express 30^(30^30) as a^(b^c) where a, b and c are integers greater than 1?
I have already drawn up a detailed solution that yields 7040 ways, but can you guys please confirm through computers?
Thanks!
EDIT: I miscalculated. It was 7041.
Last edited by phanthanhtom (2015-07-24 03:10:51)
Offline
For anyone still interested in the problem, here is my solution:
It is obvious that a has to be of the form 30^x. Hence x.(b^c) = 30^30 = 2^30 * 3^30 * 5^30. Thus b = 2^m * 3^n * 5^p such that m, n and p are not all zeroes (since b is not 1) and mc, nc and pc are all not greater than 30.
If c = 2, we have 16 choices (from 0 to 15) for each of m, n and p for a total of 16^3 - 1 choices (x would be uniquely determined, and 1 is subtracted since m, n and p cannot all be zeroes). For c = 3 we have 11^3 - 1 choices, for c = 4 we have 8^3 - 1, for c=5 7^3 - 1 etc.
When adding all those figures up to c = 30 we will get 7041.
Offline
Pages: 1