What us the remainder if 9^11 is divided by 1000 ?

↧↧↧ Detailed Video Solution to the Problem ↧↧↧




We need to find What is the remainder if 9^11 is divided by 1000?

Theory: Remainder of any number by 1000 will be equal to the last three digits of the number.

1000 = \(10^3\)

Now, if we are able to express \(9^{11}\) in terms of powers of 10 then any number with power of ≥ 3 will give zero remainder when divided by \(10^3\)

\(9^{11}\) = \((10-1)^{11}\)

Now, \((10-1)^{11}\) can be opened using Binomial theorem

=> \((10-1)^{11}\) = \(11C0 * 10^{11}(-1)^0 + 11C1 * 10^{10}(-1)^1 + 11C2 * 10^9(-1)^2 + .... + 11C8 * 10^{3}(-1)^8 + 11C9 * 10^{2}(-1)^9 + 11C10 * 10^{1}(-1)^{10} + 11C11 * 10^{0}(-1)^{11}\)

Now, all the terms starting from the first time till the \(4^{th}\) last term will have a power of 10 greater than 3
=> Their remainder by \(10^3\) will be 0

=> Our problem is reduced to what is the remainder when \(11C9 * 10^{2}(-1)^9 + 11C10 * 10^{1}(-1)^{10} + 11C11 * 10^{0}(-1)^{11}\) is divided by 1000

=> \(\frac{11*10}{2}*100*-1 + 11*10*1 + 1*1*-1\) divided by 1000
=> -5500 + 110 - 1 divided by 1000
=> -5391 divided by 1000
=> Remainder will be same as remainder of -391 by 1000
=> 1000 - 391 = 609

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Remainders

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