# finding the remainder

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• May 1st 2009, 02:25 AM
adhyeta
finding the remainder
how do we do such problems? :confused:

like...

find the remainder when $3^{37}$ is divided by 79...

or when $4^{101}$ is divided by 101...

whats the method?
• May 1st 2009, 03:33 AM
ADARSH
Quote:

Originally Posted by adhyeta
how do we do such problems? :confused:

like...

find the remainder when $3^{37}$ is divided by 79...

or when $4^{101}$ is divided by 101...

whats the method?

Easy way to solve such problem is like

since 3^4=81 we have to take value greater than 79 hence (3^4)^9*3 = 3^37

3^4/79 = 81/79 leaves remainder 2 hence 2^9 *3 / 79 = 512*3 = 1536/79 will leaves a reminder 35

I hope it is Easy solution (think of binomial expansion)
• May 1st 2009, 03:55 AM
adhyeta
Quote:

Originally Posted by ADARSH
hence (3^4)^9*3 = 3^37

dint get this
• May 1st 2009, 04:16 AM
ADARSH
Quote:

Originally Posted by adhyeta
dint get this

${(3^4)^9}*3 = 3^{37}$