how do we do such problems? like... find the remainder when $\displaystyle 3^{37}$ is divided by 79... or when $\displaystyle 4^{101}$ is divided by 101... whats the method?
Follow Math Help Forum on Facebook and Google+
Originally Posted by adhyeta how do we do such problems? like... find the remainder when $\displaystyle 3^{37}$ is divided by 79... or when $\displaystyle 4^{101}$ 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)
Originally Posted by ADARSH hence (3^4)^9*3 = 3^37 dint get this
Originally Posted by adhyeta dint get this $\displaystyle {(3^4)^9}*3 = 3^{37}$
View Tag Cloud