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

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

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23 Sep 2013, 19:47
1
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email2vm wrote:
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 and thus will be divisible by 5. The last term will be $$3^{86}$$. So we should find the remainder when $$3^{86}$$ is divided by 5.

Next, $$3^{86}=9^{43}$$. 9 in odd power has units digit of 9 hence yields the remainder of 4 upon division by 5 (9 in even power has units digit of 1 hence yields the remainder of 1 upon division by 5).

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Hope it helps.

Hi Brunel,

Either way works.

$$(45-2)^{86}$$ when expanded will have 45 in all terms except the last term $$(-2)^{86}$$. The last term of $$2^{86}$$ will be 4 (since 2 has a cyclicity of 4 so 2^86 has the same last digit as 2^2 = 4)
Also, the first term will be $$45^{86}$$ which will end with 5. All other terms will end with 0 since they have a 5 as well as a 2.

So adding all last terms, last term of $$(45-2)^{86} = ...5 + ...0 + ...0 + ...0 + ....+ ...0 + ...4$$
Last term obtained is 9 which gives 4 as remainder when divided by 5.
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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23 Sep 2013, 20:37
I did: (43)*(43^85) since 43^85 is divisible by 5, So left with 43
Where did I go wrong here?
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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23 Sep 2013, 22:13
igotthis wrote:
I did: (43)*(43^85) since 43^85 is divisible by 5, So left with 43
Where did I go wrong here?

How is 43^85 divisible by 5?
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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23 Sep 2013, 22:53
igotthis wrote:
I did: (43)*(43^85) since 43^85 is divisible by 5, So left with 43
Where did I go wrong here?

The base needs to end in a 5 to make the number divisible by 5. The multiple of 5 in the POWER has no relevance.

$$(...5)^{anything}$$ always ends in a 5.

$$(Anything)^{...5}$$ doesn't necessarily end in a 5.

e.g.
$$3^5 = 3*3*3*3*3 = 243$$
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Re: What is the remainder when 43^{86} is divided by 5?  [#permalink]

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23 Sep 2013, 23:14
VeritasPrepKarishma wrote:
Alterego wrote:
What is the remainder when $$43^{86}$$ is divided by 5?

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

Please provide a detail explanation on how you achieved the correct answer.
Thanks

First check out the post on binomial on this link: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

Now the question will take you 15 secs.

$$43^{86} = (40 + 3)^{86}$$

Since 40 is completely divisible by 5, you only have to think about $$3^{86}$$
$$3^{86} = 9^{43} = (10 - 1)^{43}$$

Again, 10 is completely divisible by 5 so we only need to worry about (-1). The remainder will be $$(-1)^{43} = -1$$ which means the remainder is 5 - 1 = 4

If you are uncomfortable with negative remainders, check this post: http://www.veritasprep.com/blog/2011/05 ... emainders/

This is nice and fantastic way to simplify the question, though new to me.

Initially I approached this question in traditional fashion.

What is the remainder when $$43^{86}$$ is divided by 5?

Rule :- The expression $$\frac{A * B * C}{M}$$ will give the same remainder as $$\frac{Ar * Br * Cr}{M}$$ where Ar, Br, Cr are the remainders of A, B, C when divided by 'M' individually.

So The remainder of $$\frac{43^{86}}{5}$$ will be the same as that of $$\frac{3^{86}}{5}$$

Remainder of $$\frac{3^1}{5}$$ = 3

Remainder of $$\frac{3^2}{5}$$ = 4

Remainder of $$\frac{3^3}{5}$$ = 2

Remainder of $$\frac{3^4}{5}$$ = 1

Remainder of $$\frac{3^5}{5}$$ = 3

The remainder pattern will repeat from 1 i.e. 34213421......so on

So the the remainder of $$\frac{3^{86}}{5}$$ will be 4.
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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24 Sep 2013, 00:09
email2vm wrote:
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 and thus will be divisible by 5. The last term will be $$3^{86}$$. So we should find the remainder when $$3^{86}$$ is divided by 5.

Next, $$3^{86}=9^{43}$$. 9 in odd power has units digit of 9 hence yields the remainder of 4 upon division by 5 (9 in even power has units digit of 1 hence yields the remainder of 1 upon division by 5).

Similar questions to practice:
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what-is-the-remainder-when-32-32-32-is-divided-by-100316.html
what-is-the-remainder-when-18-22-10-is-divided-by-99724.html

Hope it helps.

Hi Brunel,

Sure you can. 45 leaves no remainder after division by 5. so now you just have to find the remainder when (-2)^86/5
Since 86 is even, the answer will be positive. 2 has a cyclicality of 4 (i.e 2,4,8,6..2,4,6,8). so 86/4 has a remainder of 2. Since (2)^2 = 4. The number ends in 4. and the remainder when 4/5 is 4!
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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24 Sep 2013, 08:58
VeritasPrepKarishma wrote:
igotthis wrote:
I did: (43)*(43^85) since 43^85 is divisible by 5, So left with 43
Where did I go wrong here?

The base needs to end in a 5 to make the number divisible by 5. The multiple of 5 in the POWER has no relevance.

$$(...5)^{anything}$$ always ends in a 5.

$$(Anything)^{...5}$$ doesn't necessarily end in a 5.

e.g.
$$3^5 = 3*3*3*3*3 = 243$$

I understand that. But I calculated that using cyclicality. (43^85)..
1) Found the cyclicality of 3 (3,9,7,1) = 4
2) Divide the power by the cyclicality, so 85/4.. remainder is 1.
3) So the last digit of (43^85) ends in 3^1 = 3

aha! Now I get it 3/5 leaves a remainder of 3. BUT this method does not work since the other 43 is left out and creates a mess.

I'll stick with the method of (43)^86 leaves the same remainder of (3)^86 i.e 4.

Thanks Brunel and Karishma!!
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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24 Sep 2013, 20:37
igotthis wrote:
VeritasPrepKarishma wrote:
igotthis wrote:
I did: (43)*(43^85) since 43^85 is divisible by 5, So left with 43
Where did I go wrong here?

The base needs to end in a 5 to make the number divisible by 5. The multiple of 5 in the POWER has no relevance.

$$(...5)^{anything}$$ always ends in a 5.

$$(Anything)^{...5}$$ doesn't necessarily end in a 5.

e.g.
$$3^5 = 3*3*3*3*3 = 243$$

I understand that. But I calculated that using cyclicality. (43^85)..
1) Found the cyclicality of 3 (3,9,7,1) = 4
2) Divide the power by the cyclicality, so 85/4.. remainder is 1.
3) So the last digit of (43^85) ends in 3^1 = 3

aha! Now I get it 3/5 leaves a remainder of 3. BUT this method does not work since the other 43 is left out and creates a mess.

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

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25 Oct 2015, 03:10
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 and thus will be divisible by 5. The last term will be $$3^{86}$$. So we should find the remainder when $$3^{86}$$ is divided by 5.

Next, $$3^{86}=9^{43}$$. 9 in odd power has units digit of 9 hence yields the remainder of 4 upon division by 5 (9 in even power has units digit of 1 hence yields the remainder of 1 upon division by 5).

Similar questions to practice:
when-51-25-is-divided-by-13-the-remainder-obtained-is-130220.html
what-is-the-remainder-of-126493.html
what-is-the-remainder-when-32-32-32-is-divided-by-100316.html
what-is-the-remainder-when-18-22-10-is-divided-by-99724.html

Hope it helps.

Hi Bunuel, can we use cyclicity approach finding the remainder ONLY when dividing by 2, 5, or 10. Or this is a general approach unregarded how large is the divisor ? In other word could we have used the same approach if a divisor were 26 ?
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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08 Mar 2016, 03:58
VeritasPrepKarishma wrote:
idreesma wrote:
i did it the following way (just making sure that its not by luck that i got the right answer)

43^86 /5 ==> since we are looking for a remainder that is 40+3 ... 3^86 [3 repeats in the following manner
3^1=3, 3^2=9, 3^3=7, 3^4=1 [only unit digits]

86/4 ==> Remainder is 2 .. which means that unit digit is going to be 9 (9/5 ==> gives a remainder of 4 as E)

Your logic worked because we are discussing divisibility by 5 here. The last digit decides the remainder when a number is divided by 5.
Remainder when ****7 is divided by 5 will always be 2. Remainder when *****4 is divided by 5 will always be 4. This is so because every number that ends in 0 or 5 is divisible by 5 and only numbers ending in 0 or 5 are divisible by 5. Last digit works only for 2 and 5.

If you consider divisibility by say 3 or 7 etc last digit logic doesn't work so be careful.

Dear Karishma,
Please could you just quickly look through on this:

"Second, you have to know that when it comes to these kinds of questions, the only digit that matters is the UNITS DIGIT of the number."
So is it wrong OR is it true only for 2 and 5?
Can we use cyclicity approach finding the remainder ONLY when dividing by 2, 5, or 10?
Does the last digit work ONLY for 2 and 5?
If it's so, I prefer to use ONE general method in order not to mess everything up and to avoid mistakes, what do you think about it, am I right?
[Since 40 is completely divisible by 5, you only have to think about 3^86;
3^86=9^43=(10−1)^43
AND as in your post "A Tricky Question on Negative Remainders"
Question: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))
7^(11) = (8 – 1)^(11)]

- Could I also ask you what the remainder is if for example 3^77/5?
I tried to solve it this way: 3^1*3^76 / 5 = 3*(3^2)^38 / 5 = 3*9^38 / 5 = 3*(10-1)^38 / 5 --> -1^38 = +1*3 = 3 – REMAINDER.
- Is it possible to solve this question by defining the last digit:
As we know 3 has a cyclicity of 4 values in succession: 3 9 7 1 3 9 7 1 3 etc
By applying this method we need 77/4 = 1 – remainder, 1 more than 4, which is 3-->13-(5*2)=3 – ANSWER

Thank you!
Wishing u all the best
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What is the remainder when 43^86 is divided by 5?  [#permalink]

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08 Mar 2016, 23:31
2
The concept of cyclicity and units digit is useful while dealing with remainders only when the divisor is 5 or 2 or their product 10 (actually if the divisor is 2, all you need to think of is even-odd). The reason for that is that we have base 10 number system and 10 has only two prime factors 2 and 5. For more on this, check:
http://www.veritasprep.com/blog/2015/12 ... questions/
http://www.veritasprep.com/blog/2015/12 ... ns-part-2/

studentsensual wrote:

"Second, you have to know that when it comes to these kinds of questions, the only digit that matters is the UNITS DIGIT of the number."
So is it wrong OR is it true only for 2 and 5?
Can we use cyclicity approach finding the remainder ONLY when dividing by 2, 5, or 10?
Does the last digit work ONLY for 2 and 5?
If it's so, I prefer to use ONE general method in order not to mess everything up and to avoid mistakes, what do you think about it, am I right?
[Since 40 is completely divisible by 5, you only have to think about 3^86;
3^86=9^43=(10−1)^43
AND as in your post "A Tricky Question on Negative Remainders"
Question: What is the remainder when 3^(7^11) is divided by 5? (here, 3 is raised to the power (7^11))
7^(11) = (8 – 1)^(11)]

The one general method would be binomial theorem which is applicable for all divisors. But note that if you can use cyclicity in the question (if divisor is 2, 5 or 10), the solution becomes much simpler if you use cyclicity.

Quote:
- Could I also ask you what the remainder is if for example 3^77/5?
I tried to solve it this way: 3^1*3^76 / 5 = 3*(3^2)^38 / 5 = 3*9^38 / 5 = 3*(10-1)^38 / 5 --> -1^38 = +1*3 = 3 – REMAINDER.
- Is it possible to solve this question by defining the last digit:
As we know 3 has a cyclicity of 4 values in succession: 3 9 7 1 3 9 7 1 3 etc
By applying this method we need 77/4 = 1 – remainder, 1 more than 4, which is 3-->13-(5*2)=3 – ANSWER

Thank you!
Wishing u all the best

Both methods are correct. You can use binomial theorem as done in method 1 for any divisor.
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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09 Mar 2016, 21:51
Ms Karishma,

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

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17 Mar 2016, 04:56
Nice Question
Here the key is to remember that we the remainder when any integer is divided by 5 is the same as when the unit digit of the integer is divided by 5
hence as the unit digit of 43^86 is 9 (as cyclicity of 3 is 4)
hence Remainder => remainder when 9/5 => 4
hence => E
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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02 Dec 2016, 03:06
Alterego wrote:
What is the remainder when 43^86 is divided by 5?

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

Quick method.

$$43 = 3 (mod_5)$$ and we got:

$$\frac{3^{86}}{5}$$

$$5$$ is prime and $$GCD (3, 5) = 1$$ hence $$3^4 = 1 (mod_5)$$

$$\frac{(3^4)^{28}*3^2}{5}= \frac{1*9}{5}= 4 (mod_5)$$

Our remainder is $$4$$.
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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30 Dec 2016, 10:15
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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14 Jan 2017, 08:53
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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14 Jan 2017, 13:03
Alterego wrote:
What is the remainder when 43^86 is divided by 5?

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

Unit digits repeat after every 4th power.
=> 43^86=43^(84+2)

=>3^2/5=R(4)

E.

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

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14 Nov 2018, 23:37
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? &nbs [#permalink] 14 Nov 2018, 23:37

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