Finding the last two digits of a number by binomial theorem

Question

What are the last two digits of $11^{25}$ to be solved by binomial theorem like $(1+10)^{25}$? If there is any other way to solve this it would help if that is shown too.

Answer 1

$$11^{25} = (1+10)^{25} = \dbinom{25}{0} + 10\cdot\dbinom{25}{1} + \sum_{k=2}^{25}\dbinom{25}{k}\cdot10^k =$$ $$ 1 + 5\cdot10 + 2\cdot 10^2 + \sum_{k=2}^{25}\dbinom{25}{k}\cdot10^k =$$ $$\underline{1} + \underline{5}\cdot10 + 10^2\cdot \left(2+\sum_{k=2}^{25}\dbinom{25}{k}\cdot10^{k-2}\right), $$ hence the last digits are 5 and 1:

Answer 2

Here's another way. $11\times11=121$ ends in 21. $11\times21=231$ ends in 31. $11\times31=341$ ends in 41. See the pattern for the last two digits of powers of 11? Now prove that the pattern continues to hold, and then see what that tells you about $11^{25}$.

Answer 3

A short cut for finding last two digits of any power of a number that ends in 1 .

Last two digits of

$$(\ldots a1)^{\displaystyle\ldots x}$$

TENS DIGIT: unit's digit of ($x$ $\times$ a)

UNITS DIGIT : $1$

so here $$11^{25}$$ last two digits are units digit of $(1*5)$ and $1$ so $$51$$