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

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).

Answer: E.

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Hope it helps.

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).

Hi Karishma/Bunnel,

Please help with this problem:

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,


Please guide.

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.

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.

Hi Brunel,

Please help me solve when I choose (45-2)^86....i am not able to solve this way.

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.

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 my answer correct?
- 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

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.



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.



Both methods are correct. You can use binomial theorem as done in method 1 for any divisor.

Solution:

Every 43 on dividing by 5 leaves a negative remainder of -2

(Approach to get a 1 or -1 in the numerator in remainder questions to reduce investment of time)

=> (43)^86/5 ----> (-2)^86 / 5

= ((-2)^4)^21 x -2 x -2 /5

=> (1)^21 x 4 / 5 (Every 16 leaves a remainder of 1 on dividing with 5)

=> 4/5 -----> -1 as remainder = 5-1 = +4 as the positive remainder (option e)

Devmitra Sen
GMAT SME

We know to find what is the remainder when \(43^{86}\) is divided by 5

Theory: Remainder of a number by 5 is same as the unit's digit of the number

(Watch this Video to Learn How to find Remainders of Numbers by 5)

Using Above theory Remainder of \(43^{86}\) by 5 unit's digit of \(3^{86}\)

Now, Let's find the unit's digit of \(3^{86}\) first.

We can do this by finding the pattern / cycle of unit's digit of power of 3 and then generalizing it.

Unit's digit of \(3^1\) = 3
Unit's digit of \(3^2\) = 9
Unit's digit of \(3^3\) = 7
Unit's digit of \(3^4\) = 1
Unit's digit of \(3^5\) = 3

So, unit's digit of power of 3 repeats after every \(4^{th}\) number.
=> We need to divided 86 by 4 and check what is the remainder
=> 86 divided by 4 gives 2 remainder

=> \(3^{86}\) will have the same unit's digit as \(3^2\) = 9
=> Unit's digits of \(43^{86}\) = 9

But remainder of \(43^{86}\) by 5 cannot be more than 5
=> Remainder = Remainder of 9 by 5 = 4

So, Answer will be E
Hope it helps!

Watch the following video to learn the Basics of Remainders

hey Bunuel, does this approach only work when divisor = 5 (since any number with the same units digit will yield the same remainder when divided by 5)? or does this approach work in general with other divisors also (not multiples of 5)?

Check this post: https://gmatclub.com/forum/what-is-the- ... l#p1656424

For similar questions check: Units digits, exponents, remainders problems

Hope this helps.