Find the last two digits of the number 11^2016 .

I need to figure out how to hand do this and show all my work. Help pls

Start a table of the powers of 11

n 11^n
1 11
2 121
3 1331
4 14641

The last 2 digits just cycle through
11
21
31
...
91
01
11
21
...
So, 11^2016 ends in 61

Or, you might find this article on modular arithmetic useful:

https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/fast-modular-exponentiation

To find the last two digits of the number 11^2016, you can use modular arithmetic and the concept of cyclicity.

First, observe that the last two digits of any power of 11 (i.e., 11^n) repeat in cycles of four. This means that the last two digits of 11^n are the same as the last two digits of 11^(n+4), 11^(n+8), and so on.

Let's break down the calculation step by step:

1. Find the remainder when 2016 is divided by 4.
Divide 2016 by 4: 2016 รท 4 = 504 remainder 0.

The remainder is 0, indicating that 2016 is a multiple of 4.

2. Calculate 11^(2016 mod 4).
Since 2016 is a multiple of 4, 2016 mod 4 = 0. Therefore, we need to find the last two digits of 11^0.

Any non-zero number raised to the power of 0 equals 1. Therefore, 11^0 = 1.

3. Determine the last two digits of 11^0.
The last two digits of any number raised to the power of 0 are always 01. Therefore, the last two digits of 11^0 are also 01.

Hence, the last two digits of the number 11^2016 are 01.

To summarize the process:
1. Calculate the remainder when the exponent (2016) is divided by 4.
2. Find the last two digits of 11 raised to the power of the remainder obtained in step 1.
3. Determine the last two digits of the result from step 2.

In this case, since 2016 is divisible by 4, the last two digits of 11^2016 are 01.