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Remainder Theorem

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Remainder Theorem

by smishrajec » Mon Apr 09, 2012 7:02 am
Find remainder when 63^2403 is divided by 29.

Please suggest me simple ways to find the answers to questions of this type.
This question reminds me a particular type of problem, but exactly have no clear idea at all.
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by ronnie1985 » Mon Apr 09, 2012 10:20 am
The remainder theorem is lengthy here and yields R = 23

Will provide the solution as an attachment a little later
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by shubham_k » Mon Apr 09, 2012 10:37 am
ok
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by ronnie1985 » Mon Apr 09, 2012 11:10 am
Please find the solution. It is a bit lengthy
Attachments
Remainder Theorem Problem.pdf
Solution to the problem attached
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by gaurav9839 » Fri Oct 04, 2013 6:21 am
ronnie1985 wrote:The remainder theorem is lengthy here and yields R = 23

Will provide the solution as an attachment a little later


You ans is wrong, ans should be 4

see the below solution

63^2403 mod 29

since 63 and 29 are co prime and 29 is prime number

hence according tom little theorem which states a^p-1/p =1 the value will reduce to 63^23/29

now 63^23 Mod 29 = 5^23 mod 29
=5* 25^23 mod 29
4 mod 29

Hence answer is 4
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by Brent@GMATPrepNow » Fri Oct 04, 2013 6:25 am
Unless I'm missing something totally basic, this question is way outside the scope of the GMAT.

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by [email protected] » Fri Oct 04, 2013 1:13 pm
Hey All,

Brent and I agree that this question (in it's current form) is not GMAT-Like at all. While you will be asked questions about exponents and you will be asked questions about remainders, said questions will NOT look like this one.

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