↧↧↧ Detailed Video Solution to the Problem ↧↧↧
What is the remainder obtained when \(63^{25}\) is divided by 16?To solve this problem we will be using a concept called as
Binomial TheoremLearn more about Binomial Theorem
in this video.
Now, we need to break 63 into two number
- One number should be a multiple of 16 and should be close to 63 (i.e. 64)
- Other number should be a small number to make the sum or difference as 63 (i.e. -1)
=> Remainder of \(63^{25}\) by 16 = Remainder of \((64-1)^{25}\) by 16
"The reason we are doing this is because when we open \((64-1)^{25}\) this using Binomial Theorem then we will get all the terms except one term as a multiple of 64 (which also makes them a multiple of 16."
=> Remainder of all the terms by 16, except one term will be 0
Let's open \((64-1)^{25}\) using Binomial Theorem to understand this
\((64-1)^{25}\) = \(25C0 * 64^{25} * (-1)^0 + 25C1 * 64^{24} * (-1)^1 + .... + 25C24* 64^{1} * (-1)^{24} + 25C25* 64^{0} * (-1)^{25}\)
=> All terms except the last term are multiples of 64 => Their remainder by 16 will be 0
=> Our problem is reduced to what is the remainder when \(25C25* 64^{0} * (-1)^{25}\) is divided by 16
\(25C25* 64^{0} * (-1)^{25}\) = 1 * 1 * -1 = -1
=> Reminder of -1 by 16 = -1 + 16 = 15
So,
Answer will be EHope it helps!
MASTER Remainders with 2, 3, 5, 9, 10 and Binomial Theorem