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Bunuel
What is the remainder when (67^67 + 67) is divided by 68 ?

(A) 1
(B) 11
(C) 63
(D) 66
(E) 67

=67(67+1) =67*68
divisor=68
(67*68)/68
=R-67
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧



We need to find the remainder when \(67^{67} + 67\) is divided by 68?

Remainder of \(67^ {67} + 67\) by 68 = Remainder of \(67^ {67} \) by 68 + Remainder of 67 by 68

To find Remainder of \(67^ {67} \) by 68 we will be using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(68) and a big number, other part is a small number.

=> \(67^{67}\) = \((68 - 1)^{67}\)

Watch this video to MASTER BINOMIAL Theorem

Now, if we use Binomial Theorem to expand this then all the terms except the last term will be a multiple of 68
=> All terms except the last term will give remainder of 0 when divided by 68
=> Problem is reduced to what is the remainder when the last term (i.e. \(67C67 * 68^0 * (-1)^{67}\)) is divided by 68
=> Remainder of -1 is divided by 68 = 67

Remainder of \(67^{67} + 67\) by 68 = Remainder of \(67^{67}\) by 68 + Remainder of 67 by 68 = Remainder of 67 + 67 = Remainder of 68 + 66
= 0 + 66 = 66

So, Answer will be D
Hope it Helps!

Watch following video to MASTER Remainders by 2, 3, 5, 9, 10 and Binomial Theorem

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Hi rajatchopra1994, can you explain the transition between the first and the second step?
rajatchopra1994
Explanation:

(67^67)/ 68 + 67/68
(-1)^67/68 + 67/68
-1 + 67 / 68
66/68

Remainder = 66

IMO-D

Posted from my mobile device
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anurag029
That happened because of binomial theorem which I explained in my post. Once you are familiar with the concept then you can directly write remainder of 67^power by 68 as Remainder of -1^power by 68.
-1 comes by doing 67-68

Hope it helps!
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