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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Remainder Theorem
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smishrajec
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ronnie1985
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The remainder theorem is lengthy here and yields R = 23
Will provide the solution as an attachment a little later
Will provide the solution as an attachment a little later
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ronnie1985
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Please find the solution. It is a bit lengthy
- Attachments
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- Remainder Theorem Problem.pdf
- Solution to the problem attached
- (795.13 KiB) Downloaded 92 times
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gaurav9839
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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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Brent@GMATPrepNow
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Unless I'm missing something totally basic, this question is way outside the scope of the GMAT.
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Brent
Cheers,
Brent
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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.
GMAT assassins aren't born, they're made,
Rich
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.
GMAT assassins aren't born, they're made,
Rich
