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555-605 Level|   Remainders|      
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ahmed.abumera
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What is the remainder when (3^84)/26 04 Feb 2018, 02:18
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Dear Experts ,

Please let me know what is the problem in the method below;

3^84 = 9^42 = 81^21. so the question would be what is the remainder when (78+3)^21 divided by 26. since 26 is a factor of 78 that yields no remainder when dividing by 26. so we have to find the remainder when 3^21 divided by 26. Cyclisty tells us 3^21 ends with 3 in the units . so the remainder when 3 is divided by 26 is 3..
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NidSha
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What is the remainder when (3^84)/26 02 Sep 2018, 18:49
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Abhishek009
Bunuel
What is the remainder when \(\frac{(3^{84})}{26}\)

(A) 0
(B) 1
(C) 2
(D) 24
(E) 25
\(\frac{(3^{84})}{26}\)

= \(\frac{3^{3*28}}{26}\)

= \(\frac{27^{28}}{26}\)

= \(\frac{(26 + 1)^{28}}{26}\)

= \(\frac{(26^{28} + 1^{28})}{26}\)

\frac{26^{Any \ Number }}{26} , will have remainder 0
\frac{1^{Any \ Number }}{26} , will have remainder 1

So, The answer will be (B) 1
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Hi Abhishek009,

I am slightly confused = is it not wrong to say that 27^28 = (26^28 + 1^28) ??

it's like saying 2^4 = (1^4 + 1^4) which comes out to be = 2 and not 16..

Please guide?
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What is the remainder when (3^84)/26 02 Mar 2021, 22:32
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Would this same method of cyclical remainders be applicable?

3^1 = 3 --> 3/26 --> R3
3^2 = 9 --> 9/26 --> R9
3^3 = 27 --> 27/26 --> 1 R1 <---- Cycle stops here
3^4 = 81 --> 81 / 26 --> 3 R3

Given an exponent of 84 ---> 84 /3 = 26 <--- so 26 cycles of 3. Hence, the remainder is 1.
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What is the remainder when (3^84)/26 13 Mar 2021, 23:42
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Nik11
ashikaverma13
Bunuel
What is the remainder when \(\frac{(3^{84})}{26}\)

(A) 0
(B) 1
(C) 2
(D) 24
(E) 25

Hi Bunuel,

Will it be possible for you to provide an approach to this type of questions? or could you guide me to a post where you have already posted an explanation to such questions? I read the entire thread and I was unable to understand the approaches and frankly it seemed like a really time consuming approaches.
Kindly help,

thanks in advance!
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For this type of questions for me the best approach is to use cycles. Even though sometimes could take you instead of 50 sec 1:50. This simply because is the only one among the 3 that you can apply in "many" others mid-high level problems, but just my tought.

3^1 when divided by has remainder equal to 26
3^2 9
3^3 1
3^4 3 probably cycle yet finished
3^5 9 I would be sufficiently sure and I wouldn't go ahead but your choice
so the pattern is 9, 1, 3
84 is divisible by 3 hence r = 3

I want to ask to the experts what is the binomial method and if it worth to study it
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What is the remainder when (3^84)/26 01 Jul 2025, 20:15
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What is the remainder when \(\frac{(3^{84})}{26}\)

\(3^{84}\) = \(3^{3 * 24}\) = \((3^3)^{24}\) = \(27^{24}\)

To solve this problem we will be using a concept called as Binomial Theorem
MASTER Binomial Theorem in this video.

Now, we need to break 27 into two number
  • One number should be a multiple of 26 and should be close to 27 (i.e. 26)
  • Other number should be a small number to make the sum or difference as 27 (i.e. +1)
=> Remainder of \(27^{24}\) by 26 = Remainder of \((26+1)^{24}\) by 26

"The reason we are doing this is because when we open \((26+1)^{24}\) using Binomial Theorem then we will get all the terms except one term as a multiple of 26 (which also makes them a multiple of 26."
=> Remainder of all the terms by 26, except one term will be 0

Let's open \((26+1)^{24}\) using Binomial Theorem to understand this

\((26+1)^{24}\) = \(24C0 * 26^{0} * 1^{24} + 24C1 * 26^{1} * 1^{23} + .... + 24C23* 26^{23} * 1^{1} + 24C24* 26^{24} * 1^{0}\)

=> All terms except the first term are multiples of 26 => Their remainder by 26 will be 0

=> Our problem is reduced to what is the remainder when \(24C0 * 26^{0} * 1^{24}\) is divided by 26
\( 24C0 * 26^{0} * 1^{24} \) = 1 * 1 * 1 = 1
=> Reminder of 1 by 26 = 1

So, Answer will be B
Hope it helps!

MASTER Remainders with 2, 3, 5, 9, 10 and Binomial Theorem

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