Question

Carry out the fast exponentiation algorithm to evaluate 32n mod 13 for n = 0,1 and 2. These three values suce to derive a formula for 32n mod 13 for all non-negative integers n. State this formula and explain.

Answer 1

`32n\,\,mod\,\,13=6n;\,\,n=0,1,2,3,...`

Explanation:

`a\,\,mod\,\,b` refers to the quotient that is obtained on dividing
`a` by
`b`.

To find:
`32n\,\,mod\,\,13;\,\,n=0,1,2`

Solution:

For n = 0:

`32n\,\,mod\,\,13=0\,\,mod\,\,13=0`

For n = 1:

`32n\,\,mod\,\,13=32\,\,mod\,\,13=6`

For n = 2:

`32n\,\,mod\,\,13=64\,\,mod\,\,13=12`

Therefore,

`32n\,\,mod\,\,13=0\,\,mod\,\,13=0=0* 6\\32n\,\,mod\,\,13=32\,\,mod\,\,13=6=1* 6\\32n\,\,mod\,\,13=64\,\,mod\,\,13=12=2* 6`

To find: general formula for
`32n\,\,mod\,\,13`

So, as per the pattern,
`32n\,\,mod\,\,13=6n;\,\,n=0,1,2,3,...`