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# Collection of remainder problems in GMAT

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Re: Collection of remainder problems in GMAT [#permalink]

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06 Jul 2015, 04:48
girltalk wrote:
Nach0 wrote:
sandipchowdhury wrote:
2 If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ?
(1) n = 2
(2) m = 1

The Concept tested here is cycles of powers of 3.

The cycles of powers of 3 are : 3,9,7,1

St I) n = 2. This makes 3^(4*2 +2 + m) = 3^(10+m). we do not know m and hence cannot figure out the unit digit.

St II) m=1 . This makes 3^(4*n +2 + 1).
4n can be 4,8,12,16...
3^(4*n +2 + 1) will be 3^7,3^11, 3^15,3^19 ..... in each case the unit digit will be 7. SUFF
Hence B

I got this question on my GMAT Prep and got it wrong. But after looking at the question again, it seems like I got a different equation.

Mine is: 3^(4n+2)+m
Which makes a big difference!!! I answered C, which I think would be right for THIS equation, not the 3^(4n+2+m)

sorry, why would it be different? seems the same to me, as long as m=1

Can someone explain why "3^(4*n +2 + 1) will be 3^7,3^11, 3^15,3^19 ..... in each case the unit digit will be 7."

I know this post is long dead, but I'm trying my luck!

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Collection of remainder problems in GMAT [#permalink]

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06 Jul 2015, 05:31
GeeWalia wrote:

Can someone explain why "3^(4*n +2 + 1) will be 3^7,3^11, 3^15,3^19 ..... in each case the unit digit will be 7."

I know this post is long dead, but I'm trying my luck!

For unit's digit, you need to remember a couple of things:

Concept of cyclicity: cyclicity of 3 is 4. This means that unit's digit wrt 3 as the base will repeat after sets of 4

Example, 3^1 =3, 3^2=9, 3^3=27, 3^4=81, 3^5=243 etc . So you see you have unit digits as {3,9,7,1}, {3,9,7,1} etc . So the digits repeat after every set of 4. Thus, 3^15 would have the same unit digit as 3^3 as 3^15 = 3^(12+3)= same digit as 3^3 as 12 is a multiple of 4.

3^7, 3^11... will be tackled in a similar fashion.

Hope this helps.

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Re: Collection of remainder problems in GMAT [#permalink]

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18 Jul 2016, 18:21
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Collection of remainder problems in GMAT [#permalink]

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01 Jul 2017, 03:53
peraspera

if you see the "power cycle of 3" i.e 3^1, 3^2, 3^3, 3^4, 3^5, 3^6, 3^7, 3^8, 3^9, 3^10,...... the units digit is 3,9,7,1,3,9,7,1,3,9 and so on. SO you see the 3,9,7,1 pattern continues.

In the given question when m=1, we get 3^(4n+2+1)=3^(4n+3) . From the "power cycle of 3" we can see that every 4th power of 3 will end in 1 and 3 places ahead on the power cycle would have a unit digit of 7.

Remainder when a number is divided by 10 would be the units digit of that number. For the question above as we have solved the units digit would be 7 and hence remainder would be 7.

Hope this helps.

Regards,
Shreya

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Re: Collection of remainder problems in GMAT [#permalink]

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28 Oct 2017, 07:40
when the positive integer x is divided by 9 the remainder is 5 . what is the remainder when 3x is divided by 9?

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Re: Collection of remainder problems in GMAT [#permalink]

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28 Oct 2017, 07:42
SHATHY wrote:
when the positive integer x is divided by 9 the remainder is 5 . what is the remainder when 3x is divided by 9?

Discussed here: https://gmatclub.com/forum/when-the-pos ... 56032.html

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Re: Collection of remainder problems in GMAT   [#permalink] 28 Oct 2017, 07:42

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