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

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

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Updated on: 26 Feb 2019, 03:19
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What is the remainder when 43^86 is divided by 5?

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

Originally posted by Alterego on 21 Jun 2012, 15:59.
Last edited by Bunuel on 26 Feb 2019, 03:19, edited 1 time in total.
Updated.
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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21 Jun 2012, 21:13
20
21
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/
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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21 Jun 2012, 23:59
12
9
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.
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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21 Jun 2012, 16:58
16
6
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, you have to come into terms that the GMAT doesn't expect you to calculate for the value of 43^86.

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.

Always try to enumerate the powers of the said number to LOOK FOR THE PATTERN:

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81 (see it's still easy to multiply 3 from the previous digit, it's still "time-friendly")
3^5 = 243 (it's still time-friendly here)
3^6 = (now it becomes counter-productive to calculate 243*3; so what do we do then? let's just multiply the units digit by 3) = 3*3 = 9
3^7 = _ _ _ 7 (7 is the last digits, although I don't know if it's a four digit number of 5, doesn't matter)

Are you seeing the pattern? If you haven't, check out the corresponding units digit for each power

when raised to 1, the units digit is 3
raised to 2, the units digit is 9
raised to 3, the units digit is 7
raised to 4, the units digit is 1
raised to 5, the units digit is 3 <--- "the cycle begins again"
raised to 6, the units digit is 9

Now we know that raised to 6, the units digit is 9, the question says that 43 should be raised to 86 (which is equal to raised to 6, check our pattern). This means the units digit is 9

Now let's divide 9 by 5

What's the remainder? 4

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

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22 Jun 2012, 12:22
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)
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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23 Jun 2012, 02:41
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)

Yes, your approach is correct, though you could have done the second step quicker by considering 9^43 instead of 3^86 (what-is-the-remainder-when-43-86-is-divided-by-134778.html#p1098526).
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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08 Jun 2013, 05:04
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/

Hi Karishma/Bunnel,

Find the unit's digit of 47^(73^80)

Here 73^80 is number with unit's digit 1.
There fore a number ending in 7 when raised to the power a number ending in 1 - the units digit will vary
for example -
unit's digit of xy7^11 = 3,
unit's digit of xy7^21 = 7,
unit's digit of xy7^31 = 3,

however the explanation is as shown in the file attached.

I didnt get the part "If the remainder is 1, that means we will take the first term of the cycle of 7^x.

Another similar problem : 28^(43^20)

here according to me units digit will vary according to the power:

unit's digit of xy8^11 = 2,
unit's digit of xy8^21 = 8,

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

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09 Jun 2013, 21:22
2
1
cumulonimbus wrote:
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/

Hi Karishma/Bunnel,

Find the unit's digit of 47^(73^80)

Here 73^80 is number with unit's digit 1.
There fore a number ending in 7 when raised to the power a number ending in 1 - the units digit will vary
for example -
unit's digit of xy7^11 = 3,
unit's digit of xy7^21 = 7,
unit's digit of xy7^31 = 3,

however the explanation is as shown in the file attached.

I didnt get the part "If the remainder is 1, that means we will take the first term of the cycle of 7^x.

Another similar problem : 28^(43^20)

here according to me units digit will vary according to the power:

unit's digit of xy8^11 = 2,
unit's digit of xy8^21 = 8,

This question uses two related but different concepts - cyclicity and remainders

What is the last digit of $$47^5$$?

You know that the last digit is obtained by focusing on the last digit on the number i.e. 7.
7 has a cyclicity of 7, 9, 3, 1

47^1 = 47
47^2 = ...9
47^3 = ....3
47^4 = ....1
47^5 = ....7
47^6 = ....9
47^7 = .....3
47^8 = ....1
and so on

So the last digit depends on the remainder obtained when the power is divided by 4. If the power is a multiple of 4, the last digit will always be 1. If the power is 1 more than a multiple of 4 (remainder 1 when power is divided by 4), the last digit will be 7. If the power is 2 more than a multiple of 4, the last digit will be 9. If the power is 3 more than a multiple of 4, the last digit will be 3.

The concept remains the same here: $$47^x$$ where $$x = 73^{80}$$
The last digit depends on whether x is divisible by 4 or not. If not, what is the remainder?

So basically we have to take two steps:
Step I: Find the remainder when x is divided by 4.
Step II: Find the corresponding last digit for that remainder.

Step I:
$$73^{80}/4$$
$$(72+1)^{80}/4$$
(This concept is discussed above)
72 is divisible by 4.

So when $$73^{80}$$ is divided by 4, remainder is 1.

So the power of 47 is 1 more than a multiple of 4.

Step II
Knowing the cyclicity of 7, this means the last digit will be 7.
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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14 Jul 2013, 06:14
1
Here's how I did this one

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

when

$$3^1$$ divided by 5 remainder is 3
$$3^2$$ divided by 5 remainder is 4
$$3^3$$ divided by 5 remainder is (4*3) / 5 = 2
$$3^4$$ divided by 5 remainder is ( 2*3)/ 5 = 1
$$3^5$$ divided by 5 remainder is 3

So the repeating block is {3,4,2,1}

when we divide 86 by 4 the remainder is 2

Hence the remainder for this problem is 4.
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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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14 Jul 2013, 08:48
Alterego wrote:
What is the remainder when 43^86 is divided by 5?

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

Remainder$$[\frac{43^{86}}{5}] \to Rem[\frac{3^{86}}{5}]$$;Because the remainder when 43 is divided by 5 is 3,

Also,$$Rem[\frac{3^{86}}{5}]$$ =$$Rem[\frac{3^{4*21}*3^2}{5}]$$ = $$Rem[\frac{81^{21}*3^2}{5}]$$ = $$Rem[\frac{1^{21}*9}{5}]$$ = $$Rem[\frac{9}{5}]$$ = 4.

Note: It is desirable to modify the expression in such a way that we get a remainder of 1, when divided by 5. Thus, we take out $$3^4$$ from the given expression, which results in 81.

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

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23 Sep 2013, 11:47
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 Brunel,

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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
1
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:
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 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, 22:53
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, 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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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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Re: What is the remainder when 43^86 is divided by 5?  [#permalink]

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08 Mar 2016, 23:31
2
1
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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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 $$4$$.
Re: What is the remainder when 43^86 is divided by 5?   [#permalink] 02 Dec 2016, 03:06

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