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

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Intern
Joined: 07 Dec 2016
Posts: 34
What is the remainder when 5^68 is divided by 7?  [#permalink]

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14 Apr 2017, 04:26
3
25
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Difficulty:

75% (hard)

Question Stats:

50% (01:56) correct 50% (01:55) wrong based on 325 sessions

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

A) 1
B) 2
C) 3
D) 4
E) 6
Math Expert
Joined: 02 Aug 2009
Posts: 8757
Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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14 Apr 2017, 06:47
3
5
amathews wrote:
What is the remainder when 5^68 is divided by 7?

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

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
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Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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14 Apr 2017, 06:45
5
5
amathews wrote:
What is the remainder when 5^68 is divided by 7?

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

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

$$\frac{2^1}{7}$$= 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
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Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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11 May 2017, 11:30
Hi,

Can anyone please tell me why we did (7-2) at the beginning and not (8-3) or some other number? What is the concept behind this? Would be greatly appreciated if someone could please explain me the reasoning and concept behind this trick!

Thanks!
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Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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12 May 2017, 00:19
1
1
amathews wrote:
What is the remainder when 5^68 is divided by 7?

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

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

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Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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12 May 2017, 00:23
csaluja wrote:
Hi,

Can anyone please tell me why we did (7-2) at the beginning and not (8-3) or some other number? What is the concept behind this? Would be greatly appreciated if someone could please explain me the reasoning and concept behind this trick!

Thanks!

The reason is that we need to find the remainder when dividing by 7. Hence 5=7-2.

Also, when we extract the expression $$(7-2)^n$$ we will have an expression like $$7^k \times A + (-2)^n$$ with $$k$$ is an positive integer.

Note that $$7^k \times A$$ is divisible by 7, so we simply find the remainder of $$(-2)^n$$ when dividing by 7.

Hence, we won't do like this way 5=8-3 since 8 isn't divisible by 7 and this method makes the problem more complex.
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Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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12 May 2017, 02:16
amathews wrote:
What is the remainder when 5^68 is divided by 7?

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

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.
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Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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12 May 2017, 02:27
shashankism wrote:
amathews wrote:
What is the remainder when 5^68 is divided by 7?

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

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.

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
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Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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15 May 2017, 09:21
1
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
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Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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12 Sep 2018, 06:54
chetan2u wrote:
amathews wrote:
What is the remainder when 5^68 is divided by 7?

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

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

Hi Chetan.....can you tel me if the method below will also work?

5^68/7 equals (125^22 x 5^2)/7
thus we get a remainder of -1 for 125^22, which means:
(-1)^22 x 25 -> 1 x 25 -> 25 -> 25/7 -> remainder is 4.

thanks
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Posts: 8757
Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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12 Sep 2018, 06:58
Mansoor50 wrote:
chetan2u wrote:
amathews wrote:
What is the remainder when 5^68 is divided by 7?

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

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

Hi Chetan.....can you tel me if the method below will also work?

5^68/7 equals (125^22 x 5^2)/7
thus we get a remainder of -1 for 125^22, which means:
(-1)^22 x 25 -> 1 x 25 -> 25 -> 25/7 -> remainder is 4.

thanks

Yes Mansoor, you are absolutely correct with your approach
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What is the remainder when 5^68 is divided by 7?  [#permalink]

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Updated on: 09 Oct 2019, 18:34
1
amathews wrote:
What is the remainder when 5^68 is divided by 7?

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

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.

Hope this helps.

All the best!
Experts' Global Team
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Originally posted by Maxximus on 16 Sep 2018, 11:45.
Last edited by Maxximus on 09 Oct 2019, 18:34, edited 1 time in total.
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Posts: 14
Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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23 Sep 2019, 06:51
chetan2u wrote:
amathews wrote:
What is the remainder when 5^68 is divided by 7?

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

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

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.
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Re: What is the remainder when 5^68 is divided by 7?  [#permalink]

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29 Mar 2020, 01:14
amathews wrote:
What is the remainder when 5^68 is divided by 7?

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

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
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Re: What is the remainder when 5^68 is divided by 7?   [#permalink] 29 Mar 2020, 01:14