We need to find the remainder of \(132^5 − 2*132^4 + 6*132^3 −3*132\) by 65We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(65 or 130) and a big number, other part is a small number.=> \(132^{5}\) = \((130 + 2)^{5}\)
Watch this video to MASTER BINOMIAL TheoremNow, when we expand this expression then all the terms except the last term will be a multiple of 130 or 65.=> All terms except the last term will give 0 as remainder then divided by 65
=> Problem is reduced to what is the remainder when the last term is divided by 65
=> What is the remainder when \(5C5 * 130^0 * 2^{5}\) is divided by 65 = Remainder of \(2^{5}\) by 65 = 32
=> Remainder of \(132^5 \) by 65 = remainder of 32 by 65 = 32
Remainder of 2*132^4 by 65 = Remainder of 2*2^4 by 65 = Remainder of 32 by 65 = 32
Remainder of 6*132^3 by 65 = Remainder of 6*2^3 by 65 = Remainder of 48 by 65 = 48
Remainder of 3*132 by 65 = Remainder of 3*(130 + 2) by 65 = Remainder of 3*2 by 65 = 6
=> Total Remainder = 32 - 32 + 48 - 6 = 42
So,
Answer will be BHope it Helps!
Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem