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Question Stats:
0% (00:00) correct
100% (00:25) wrong based on 2 sessions
25^26^27 divided by 7 gives remainder?? a)3 b)4 c)2 d)5
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Manager
Status: Meh, I can't take the GMAT before 2017.
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This is not a GMAT question.
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Hit kudos if my post helps you. You may send me a PM if you have any doubts about my solution or GMAT problems in general.
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Senior Manager
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can someone provide a approach to solve this problem
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Manager
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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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Senior Manager
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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#p651942Formula 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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Intern
Joined: 03 Oct 2011
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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#p651942Formula 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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Intern
Joined: 03 Oct 2011
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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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