Is (expr) a natural number?

Is (19^92 - 91^29) / 90 a natural number or not?
Motivate well...

19^92 (mod 10)
=-1^92 (mod 10)
= 1 (mod 10
19^92 (mod 9)
=1 (mod 9)
Therefore by CRT 19^92=1 (mod 90)
Clearly 91^29=1 (mod 90), therefore the numerator is 0 (mod 90), yes it is an integer. Let me work out whether it is positive...

Originally posted by FabianFnas
Is [b](19^92 - 91^29) / 90 a natural number or not?
Motivate well...[/b]

We only need to know modulos for 2, 5, and 9, right? If all 3 are zero for the numerator, then we have a natural number.

2 - Odd minus odd. ZERO
5 - Cycles in 4, 4^4 - 1^1 = 15, ZERO
9 - Cycles in 6, 1^2 - 1^5 = 0, ZERO

Since all 3 are zero, the difference is divisible by 90..

(Ignore this math) 3*92 = 276, 4.8*29 = 139.2, 276 - 48 = 228, estimate value has 22-23 digits in it and is positive..