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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