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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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Alterego wrote:
What is the remainder when 43^86 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

$$\frac{43}{5}$$ = $$Remainer$$ $$3$$

We know, $$\frac{3^4}{5}$$ = $$Remainer$$ $$1$$

Now, $$3^{86} = 3^{84}*3^2$$

$$\frac{3^2}{5}$$ = $$Remainer$$ $$4$$

Hence, the correct answer will be (E) 4

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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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Bunuel wrote:
What is the remainder when 43^86 is divided by 5?
A. 0
B. 1
C. 2
D. 3
E. 4

Notice that $$43^{86}=(40+3)^{86}$$. Now, if we expand this expression, all terms but the last one will have 40 as multiple a

V.Good explanation as always but I would like to add some explanation for the above statement:

Expanding (40+3) ^ 86 means simply that you do (40+3) (40+3) (40+3) ...etc 86 times. Simply if you do all the multiplication you will end up with terms that are mul(40) apart from one which will be mul(3)... perhaps this doesn't make sense much but check the following example:

E.g. (40+3)^2 = (40+3)*(40+3) = 40*40 + 40*3 + 3*40 + 3*3 .... Indeed all the terms of this equation are multiples of 40 apart from the last term 3*3

This approach applies to any powers as for example (40+3)^3 and so on.
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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Alterego wrote:
What is the remainder when 43^86 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

[43^86]/5
= [3^86/5]
= [9^43/5]
= [-1^43/5]
= 4

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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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I tried something slightly different. Would really appreciate any input on whether I'm correct on this train of thought.

\frac{43^86}{5} can be written as \frac{[45 + (-2)]^86}{5}. Since 45 can be fully divided with 5, we ignore it.

So the question stands as: what is the remainder of 2^86/5. Following the cyclicity rule, we know that 2^86 has a units digit of ..4.

Every number that ends in 4 and is divided by 5, gives a remainder of 4, so answer is E.
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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It does work. $$43^{85}$$ ends in a 3. You have a 43 outside.

You get 43*(..........3)
The last digit here will be product of the last digits 3*3 = 9
When you divide it by 5, the remainder will be 4.

The only thing is, you dont need to take a 43 out. What did you achieve by doing that?
Using cyclicity, $$43^{86}$$, divide 86 by 4 to get 2. So last digit of $$43^{86}$$ ends in 3^2 = 9.
When you divide it by 5, the remainder will be 4.

Can we always use the cyclicity method to do these type of remainder questions? If yes, why doesn't it work for the below:

(5^68)/7 (original link: https://gmatclub.com/forum/what-is-the- ... fl=similar)

I mean 5 has a cyclicity of 1. So the last digit of 5^68 will be a 5. Now if you divide 5 by 7 (so 5/7) you get a remainder of 5. However this isn't the correct answer? Why not?
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What is the remainder when 43^86 is divided by 5?  [#permalink]

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Alterego wrote:
What is the remainder when 43^86 is divided by 5?

A. 0
B. 1
C. 2
D. 3
E. 4

Asked: What is the remainder when 43^86 is divided by 5?

Remainder when 5 divides$$43^{86}= (40+3)^{86} = 3^{86} = 81^{21} * 3^2 = 9 = 4$$

IMO E

Originally posted by Kinshook on 19 Aug 2019, 09:38.
Last edited by Kinshook on 21 Aug 2019, 07:21, edited 2 times in total.
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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jamalabdullah100 wrote:
It does work. $$43^{85}$$ ends in a 3. You have a 43 outside.

You get 43*(..........3)
The last digit here will be product of the last digits 3*3 = 9
When you divide it by 5, the remainder will be 4.

The only thing is, you dont need to take a 43 out. What did you achieve by doing that?
Using cyclicity, $$43^{86}$$, divide 86 by 4 to get 2. So last digit of $$43^{86}$$ ends in 3^2 = 9.
When you divide it by 5, the remainder will be 4.

Can we always use the cyclicity method to do these type of remainder questions? If yes, why doesn't it work for the below:

(5^68)/7 (original link: https://gmatclub.com/forum/what-is-the- ... fl=similar)

I mean 5 has a cyclicity of 1. So the last digit of 5^68 will be a 5. Now if you divide 5 by 7 (so 5/7) you get a remainder of 5. However this isn't the correct answer? Why not?

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Joined: 16 Oct 2010
Posts: 10226
Location: Pune, India
Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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jamalabdullah100 wrote:
jamalabdullah100 wrote:
It does work. $$43^{85}$$ ends in a 3. You have a 43 outside.

You get 43*(..........3)
The last digit here will be product of the last digits 3*3 = 9
When you divide it by 5, the remainder will be 4.

The only thing is, you dont need to take a 43 out. What did you achieve by doing that?
Using cyclicity, $$43^{86}$$, divide 86 by 4 to get 2. So last digit of $$43^{86}$$ ends in 3^2 = 9.
When you divide it by 5, the remainder will be 4.

Can we always use the cyclicity method to do these type of remainder questions? If yes, why doesn't it work for the below:

(5^68)/7 (original link: https://gmatclub.com/forum/what-is-the- ... fl=similar)

I mean 5 has a cyclicity of 1. So the last digit of 5^68 will be a 5. Now if you divide 5 by 7 (so 5/7) you get a remainder of 5. However this isn't the correct answer? Why not?

Units digit gives the remainder only when the divisor is one of 2/5/10.

Check out these posts:
https://www.veritasprep.com/blog/2015/1 ... questions/
https://www.veritasprep.com/blog/2015/1 ... ns-part-2/
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Veritas Prep GMAT Instructor Re: What is the remainder when 43^86 is divided by 5?   [#permalink] 21 Aug 2019, 21:09

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