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remainder

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remainder [#permalink] New post 01 Jul 2013, 06:13
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Find the remainder when 54^124 is divided by 17.

a)4
b)5
c)13
d)15



Please provide the solution with explanation.
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Re: remainder [#permalink] New post 01 Jul 2013, 19:31
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SAMHITAC wrote:
Find the remainder when 54^124 is divided by 17.

a)4
b)5
c)13
d)15



Please provide the solution with explanation.


With such big numbers, the use of binomial becomes efficient. Check this post for concepts of binomial theorem for remainder questions: http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

\(54^{124} = (51 + 3)^{124}\)
51 is divisible by 17 so we only have to worry about the remainder when \(3^{124}\) is divided by 17. Look for a power of 3 which is close to 17 - we don't see a power of 3 which is one more or one less than 17 but we do have a power of 2 which is one less than 17. So if we get a power of 3 that is 1 or 2 less or more than a multiple of 17, that will work too.

\(3^{124} = 9^{62} = ( 17 - 8)^{62}\)

So now we have to worry about \((-8)^{62}\) only

\((-8)^{62} = 2^{186} = 2^2 * 16^{46} = 4*(17 - 1)^{46}\)

Since \(-1^{46} = 1\), remainder will be 4*1 = 4
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Re: remainder   [#permalink] 01 Jul 2013, 19:31
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