how do we do such problems? :confused:

like...

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

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

whats the method?

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- May 1st 2009, 01:25 AMadhyetafinding the remainder
how do we do such problems? :confused:

like...

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

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

whats the method? - May 1st 2009, 02:33 AMADARSH

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, 02:55 AMadhyeta
- May 1st 2009, 03:16 AMADARSH