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ram
Joined: 07 Oct 2007 Posts: 7
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Posted: Wed Nov 07, 2007 8:23 am Post subject: What is the remainder? |
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What is the remainder when 5^23 is divided by 17?
Tried remainder theorem method, but it becomes too complicated. Help solicited. Thanks |
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2iim Forum Moderator
Joined: 10 Oct 2007 Posts: 8
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Posted: Sat Nov 17, 2007 12:18 pm Post subject: 10 is the remainder. |
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one approach to solve this question is as follows
remainder when 5 divided by 17 is 5
remainder when 5^2 divided by 17 is 8
remainder when 5^4 divided by 17 is 8*8 = 64 = 13
remainder when 5^8 divided by 17 is 13*13 = 169 = 16
remainder when 5^16 divided by 17 is 5 = 16*16 = 256 = 1.
5^23 can be written as (5^16) * (5^4) * (5^2) * 5
Therefore, when 5^23 is divided by 17 will be the product of the remainders of 5^16, 5^4, 5^2, and 5 divided by 17.
The remainders are 1 * 13 * 8 * 5 = 520.
As 520 is greater than 17, the remainder when 520 is divided by 17 will be the remainder.
i.e., 10 is the remainder. |
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ram
Joined: 07 Oct 2007 Posts: 7
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Posted: Wed Nov 28, 2007 5:47 pm Post subject: |
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thanks for the explanation
is there any other shorter method? |
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