chetan2u wrote:

rencsee wrote:

If n and m are positive integers, what is the last digit of \(83^{56m} + (10^2 – 1)^{64n}\)?

A. 0

B. 1

C. 2

D. 3

E. 9

\(83^{56m} + (10^2 – 1)^{64n}\)...

Now

\(83^{56m}\) will have the same units digit as 3^4 as 56m=4*14m, so 1

\((10^2 – 1)^{64n}=(99)^{64n}\) will have same units digit as 9^2 so 1

Ans 1+1=2

C

I couldnot understand your solution.

If the power of the digit of 9 is odd, then the unit digit is 9 and if the power is even, then unit digit is even.

So what if n was equal to any odd number? Then the unit digit will 9 and the resulting answer will be 0?

Where am I incorrect in saying this?