Afc0892 wrote:

What is the remainder when \(5^{91}\) is divided by 91?

1) 5

2) 8

3) 13

4) 34

5) 47

I doubt such a question will come in actuals..

Otherwise..

1) as per remainder theorem

\(91=7*13\) so LCM of (7-1,13-1) or (6,12) is 12

So \(5^{91}\) will give the same remainder as \(5^7\), since 91/12 gives 7 as remainder.

\(5^7=(5^3)^2*5=(125)^2*5\)

125 will leave 125-91=34 as remainder

So \(34*34*5=34*170\)

170 will leave a remainder of -12

\(34*(-12)= 68*(-6)=-23*-6=138\)

So remainder=138-91=47

2) if you do not know the theorem you will have to get the terms closer to multiple of 91So \(5^{91}=(5^4)^{22}*5^3=625^{22}*5^3\)

91*7=637, so remainder is \(625-637=(-12)\)

So new term is \((-12)^{22}*5^3\)

As even power, it will be same as \(12^{22}*5^3=4^{22}*3^{22}*5^3\)

Also \(4^3=64\) and remainder is 64-91=-27

So \((4^3)^7*4*3^{22}*5^3=(-27)^7*3^{22}*5^3=(-3^3)^7*3^{22}*5^3=-(3^{43}*5^3*4)\)

Now \(3^6=729\) and 91*8=728 so remainder is 1

\(-(3^6)^7*3*5^3*4=-(1^7*3*5^3*4)=-1*3*5*100=-1*3*5*9=-135\)

So POSITIVE remainder will be \(91*2-135=182-135=47\)

E

you could have used this too to simplify

Now \(2^7*5=128*5=640\) remainder will be 640-637=3

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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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