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remainder

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remainder [#permalink]

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New post 14 Oct 2011, 06:30
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25^26^27 divided by 7 gives remainder??
a)3
b)4
c)2
d)5
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Re: remainder [#permalink]

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New post 14 Oct 2011, 07:05
This is not a GMAT question.
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Re: remainder [#permalink]

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New post 14 Oct 2011, 07:51
can someone provide a approach to solve this problem
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Re: remainder [#permalink]

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New post 24 Oct 2011, 20:37
This looks like a CAT question with only 4 answer choices.

Re-phrase the question as find remainder of 25^x/7.

Now check for periodicity of 25^x/7.

When x=1, remainder=4
When x=2, remainder=2
When x=3, remainder=1

When x=4, remainder=4
...

So periodicity of 25^x is 3. We need to reduce x (i.e. 26^27) to fall into this periodicity.

Find remainder of 26^27/3:

26. 26^26/3
=> 26.2^26/3
=> 26.2^2.2^24/3
=> 26.4.(2^3)^8/3
=> 26.4.1/3
=> 104/3 = 2

Therefore, we need to find 25^2/7's remainder which is, by our periodicity map, 2. (C).
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Re: remainder [#permalink]

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New post 25 Oct 2011, 09:19
sonugupta wrote:
25^26^27 divided by 7 gives remainder??
a)3
b)4
c)2
d)5



its a very easy question if you have read " Compilation of tips and tricks to deal with remainders"
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Formula is [(25/7)(26/7)(27/7)] divided by 7

25/7 = 3 remainder, 26/7 = 4 remainder, 27/7 = 5 remainder

3.4.5/7 = 60/7 = 4 remainder
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Re: remainder [#permalink]

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New post 25 Oct 2011, 09:30
shrive555 wrote:
sonugupta wrote:
25^26^27 divided by 7 gives remainder??
a)3
b)4
c)2
d)5



its a very easy question if you have read " Compilation of tips and tricks to deal with remainders"
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Formula is [(25/7)(26/7)(27/7)] divided by 7

25/7 = 3 remainder, 26/7 = 4 remainder, 27/7 = 5 remainder

3.4.5/7 = 60/7 = 4 remainder



Don't you think 25/7 will leave a remainder 4 ?? 26/7-5 and 27/7-6 ????
3*7=21 and 25-21=4 not 3
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Re: remainder [#permalink]

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New post 25 Oct 2011, 09:44
Ans should be b ie 4.

(25^16^27)/7. In this case we follow an approach from top to down. Consider 27 first. 27/7 leaves a remainder 6. Then consider 26^6 (in place of 26^27), -> (26*26*26*26*26*26)/7. Each 26 will leave a remainder 5 as 26/7 leaves a remainder 5. so (5*5*5*5*5*5)/7 =(25*25*25)/7 and each 25 gives a remainder of 4. Leaving us with (4*4*4)/7= (16*4)/7=2*4=8/7=1 now (25^1)/7 will give 4.
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Re: remainder [#permalink]

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New post 27 Feb 2016, 03:32
The answer is B. The remainder is 2.

25 = (21 +4) = (21+2^2)
25 ^26^27 = (21 +4)^26^27
Since 21 is divided by 7 then the remainder should be the remainder of 4 ^26^27 divided by 7.
4^26^27 = 2 ^2^26^27.

We have:
2^1 :7, r = 2.
2^2 :7, r =4.
2^3:7, r=1.
2^4:7, r= 2.
....
Therefore the cycle is 3. we have A= 2 ^2^26^27 = 2^(2*26*27).
Since 2*26*27 is divided by 3 without remainder. Therefore, the remainder of A is 2.
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Re: remainder   [#permalink] 27 Feb 2016, 03:32
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