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

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.

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]
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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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]
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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/05 ... emainders/




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.
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Re: What is the remainder when 43^86 is divided by 5? [#permalink]
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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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/05 ... emainders/




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

Answer: E.

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,

Please help me solve when I choose (45-2)^86....i am not able to solve this way.
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Re: What is the remainder when 43^86 is divided by 5? [#permalink]
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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).

Answer: E.

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,

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

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 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


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

Originally posted by KarishmaB on 09 Mar 2016, 00:31.
Last edited by KarishmaB on 21 Dec 2023, 08:13, edited 1 time in total.
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Re: What is the remainder when 43^86 is divided by 5? [#permalink]
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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)

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

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


Yes, your approach is correct, though you could have done the second step quicker by considering 9^43 instead of 3^86 (https://gmatclub.com/forum/what-is-the- ... l#p1098526).


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)?
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Re: What is the remainder when 43^86 is divided by 5? [#permalink]
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tickledpink001 wrote:
Bunuel 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)


Yes, your approach is correct, though you could have done the second step quicker by considering 9^43 instead of 3^86 (https://gmatclub.com/forum/what-is-the- ... l#p1098526).


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

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