What is the remainder when 5^68 is divided by 7?

Question

A) 1
B) 2
C) 3
D) 4
E) 6

chetan2u · Accepted Answer

Hi..

Make use of expansion ...
 5^{68}=(7-2)^{68} 
When you expand the equation, all terms will be div by 7 except 
 (-2)^{68}=2^{68}=(2^3)^{22}*2^2=8^{22}*4=(7-1)^{22}*4 ..
In (7-1)^22 all terms will be div by 7 except (-1)^22, which is same as 1..
So remainder is 1*4=4
D

Abhishek009 · Answer

\frac{5^{68}}{7} = \frac{( 7 - 2 )^{68}}{7}

\frac{2^1}{7} = Remainder 2
 \frac{2^2}{7} = Remainder 4
 \frac{2^3}{7} = Remainder 1

\frac{2^4}{7} = Remainder 2
 \frac{2^5}{7} = Remainder 4
 \frac{2^6}{7} = Remainder 1

So, We have a pattern here, the cyclicity is of 3.

\frac{68}{3}  = Remainder is 2

Thus, the remainder corresponding to the value will be 4, answer must be (D) 4

broall · Answer

$\frac{5^{68}}{7}=\frac{(-2)^{68}}{7}=\frac{2^{68}}{7}=\frac{(2^3)^{22}\times 2^2}{7}
=\frac{8^{22} \times 2^2}{7}=\frac{1\times 2^2}{7}=4$

The answer is D.

shashankism · Answer

We can do it by expanding the number .. {} remainder value in the below explanation
{5^68 / 7}= {25^34 /7} = {4^34 /7} = {16^17 /7 } ={2^17/7} = {((2^3)^5 * 2^2) /7} = {((8)^5 * 2^2) /7} = 1 * 4 = 4

So Remainder is 4.

shashankism · Answer

This can also be solved by following method .. Considering -ve values in expansion.

{5^68 / 7} = {-2^68 /7} = {2^68 /7 } = {8^22*2^2/7} = { 4/7 } = 4

msrn28 · Answer

The expansion can be written as (126-1)^22 *5^2 divided by 7.

This approach would be very simple and easy to solve this type of questions;

Take 5^3 = 125 and restructure the question like this : (5^3)^22 * 5^2 divided by 7.

The main idea is to expand the numerator with a number close to 7 or multiples of 7 by 1.

If I take 5^3 =125 = 126-1. Here 126 is divisible by 7 or a multiple of 7.

Now it becomes very simple. If we expand this (126-1)^22 using binomial expansion, the remainder term would be (-1)^22, which is 1.

So we left with 5^2 divided by 7 which is 25/7 and the remainder is 4.

Hence always try to expand the numerator such that we convert a big part of the problem into 1 and then find the remainder on the remaining part.

Thanks

Maxximus · Answer

Question is 5^68%7 = ?

This can be written as (125)^22 x 5^2

Now find 125^22 & 7 and 5^2%7

125% 7 = 6 or -1
So, 125^22%7 = -1^22 = 1

5^2%7
25%7 = 4

Hence, the remainder is 1x4 = 4

D.

To understand a smart approach for solving such questions in 1-minute, please watch the following video; the concept has been explained in ~3 minutes.

https://www.youtube.com/watch?v=_f66rdkrs04

Hope this helps.

All the best!
Experts' Global Team

ypetrunina · Answer

Hello. Could you please tell me how this approach called? I still don't understand it. I thought we can use cyclicity of a unit digit to find out the remainder. 5 has 1 cyclicity and 5 to any power will end with 5. Therefore, 5/7 where remainder should be 5. Please help me to understand. Thank you.

Kinshook · Answer

Asked: What is the remainder when 5^68 is divided by 7?

5^68mod7 = (-2)^68mod7 = 2^68mod7 = 8^22*2^2mod7 = 4mod7 = 4

IMO D

samc88 · Answer

All great ways - I did it a slightly different way by trying to find a multiple of 5 that was +/- 1 away from a multiple of 7

5^3 = 125

126 is a multiple of 7

so you can break down 5^68 = (5^2)*(5^3)^22 = (5^2)*(125)^22

This can then be rewritten as : (5^2)*(126-1)^22

So reduces down because of the binomial effect to (5^2)*(-1)^22 = (5^2)

Effectively you need to find the remainder when 25 is div by 7.... so 4.

Nonbiryani · Answer

Hi, why is the below method not working?

5^68 / 7

So if we take remainders of each power of 5 divided by 7,
So, 5^1/7 = 5
5^2 / 7 = 4
5^3 / 7 = 6
5^4 / 7 = 2
.....And then this cycle continues

Now we have 5^68/7, and 68 is divisible by 4, hence by this cyclicity, the remainder would be 2

Bunuel · Answer

That's not correct. The cycle does not continue as you described:
 
The remainder of 5^1/7 is 5;
The remainder of 5^2/7 is 4;
The remainder of 5^3/7 is 6;
The remainder of 5^4/7 is 2;
The remainder of 5^5/7 is 3;
The remainder of 5^6/7 is 1;
The remainder of 5^8/7 is 4;
The remainder of 5^9/7 is 6;
The remainder of 5^10/7 is 2;
The remainder of 5^11/7 is 3;
The remainder of 5^12/7 is 1;
 
Then, the cycle repeats:
 
The remainder of 5^13/7 is 5;
The remainder of 5^14/7 is 4;
The remainder of 5^15/7 is 6;
The remainder of 5^16/7 is 2;
...
 
Hence, the cycle is {5, 4, 6, 2, 3, 1, 4, 6, 2, 3, 1}, so there are 11 different remainders. 68 divided by 11 gives a remainder of 2, so the remainder corresponds to the 2nd number in the pattern, which is 4.

However, this is not the best method to solve this question.

BrushMyQuant · Answer

We need to find the remainder when  5^{68}  is divided by 7? 
 
 We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(7) and a big number, other part is a small number.

=>  5^{68}  =  (7-2)^{68}

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 7
=> All terms except the last term will give remainder of 0 when divided by 7 
=> Problem is reduced to what is the remainder when the last term (i.e. 68C68 * 7^0 * (-2)^68) is divided by 7
=> Remainder of 2^68 is divided by 7

To solve this problem we need to find the cycle of remainder of power of 2 when divided by 7

Remainder of  2^1  (=2) by 7 = 2
Remainder of  2^2  (=4) by 7 = 4 
Remainder of  2^3  (=8) by 7 = 1
Remainder of  2^4  (=16) by 7 =   2   
Remainder of  2^5  (=32) by 7 =   4  
Remainder of  2^6  (=64) by 7 =   1

=> Cycle is 3

Remainder of 68 by 3 = 2
=> Remainder of  2^{68}  by 7 = Remainder of  2^{2}  by 7 = 4

So,  Answer will be D 
Hope it Helps!

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

https://www.youtube.com/watch?v=h1HTEhmUb1w

HRX273 · Answer

hey, can anyone help me to deal with exponents problem..

Bunuel · Answer

Exponents and Roots of Numbers

Theory: 
 
  Exponents and Roots on the GMAT: Tips and hints  
  Cyclicity and the remainders on the GMAT

Questions 
 
  Units digits, exponents, remainders problems  
  Tough and tricky DS exponents and roots questions with detailed solutions  
  Tough and tricky PS exponents and roots questions with detailed solutions  
  Exponents PS Questions  
  Roots PS Questions

Check below for more: 
 
  ALL YOU NEED FOR QUANT ! ! !  
  Ultimate GMAT Quantitative Megathread

Hope this helps.

Adit_ · Answer

Well its good to learn some remainder theorem concepts here that will come super handy and you can solve really really fast.

Fermat's little theorem states that 5^6 mod 7 would give remainder as 1.
Thus 5^66 mod 7 is also 1
5^2 mod 7 = 4
Answer = 4

yashsharma1 · Answer

I tool a slightly longer approach as mentioned below. If some expert ( Bunuel  or @chetan2u) can check it so that I am clear if this is the correct approach and it is not misleading to others as well, it would be great.

Remainder (R) of (5^68)/7 = R (25^34)/7
Now, when I divide 25 by 7, the remainder is 4 and hence
R (25^34)/7 = R (4^34)/7 = R (16^17)/7
When I divide 16 by 7, the remainder is 2 and hence
R (16^17)/7 = R (2^17)/7
I can write 2^17 as 8^5 x 4. So it becomes R (8^5 x 4)/7
When 8 in this expression is divided by 7, each time the remainder is 1 and hence
R (8^5 x 4)/7 = R (1^5 x 4)/7
We can ignore the 1 here and rewrite this as 
R (1^5 x 4)/7 = R 4/7 and hence the remainder will be 4

Adit_ · Answer

Not an expert, but that method is absolutely perfect. If you truly want a shortcut however, checkout Fermat's theorem.
Using that you can quickly conclude that 5^6 mod 7 = 1 thus 5^66 mod 7 = 1 and it reduces to 5^2 mod 7 = 4.

chetan2u · Answer

You are absolutely correct. The concept in this or any similar method is that the remainder can be multiplied and added.
For example..
1) Remainder when 8+9 is divided by 7... 1+2
2) Remainder when 8*9 is divided by 7... 1*2

What is the remainder when 5^68 is divided by 7?

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GMAT Problem Solving (PS) Questions

What is the remainder when 5^68 is divided by 7?

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