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What is the remainder when 9^381 is divided by 5? How do you

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What is the remainder when 9^381 is divided by 5? How do you [#permalink]

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New post 30 Oct 2008, 23:07
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the remainder when 9^381 is divided by 5?

How do you do such problems?

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Re: Remainder [#permalink]

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New post 30 Oct 2008, 23:28
udribat wrote:
What is the remainder when 9^381 is divided by 5?

How do you do such problems?


We use the "Remainder-Pattern" method.

remainder of (9^1)/5 = 9/5 ===> 4
remainder of (9^2)/5 = 81/5 ===> 1
remainder of (9^3)/5 = 729/5 ===> 4
remainder of (9^4)/5 = 6561/5 ===> 1

We see a pattern emerge ....
when x= odd the remainder of (9^x)/5 ===> 4
when x=even the remainder of (9^x)/5 ===> 1

in the question x=381 (it is odd)

remainder is 4.
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"You have to find it. No one else can find it for you." - Bjorn Borg

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Re: Remainder [#permalink]

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New post 30 Oct 2008, 23:30
9^1 ends with 9
9^2 ends with 1.
9^3 ends with 9.
9^4 ends with 1.

Thus, 9^380 (even power) will end with 1 and 9^381 (odd power) will end with 9 and hence the remainder will be 4.

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Re: Remainder [#permalink]

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New post 30 Oct 2008, 23:30
Remainder is 4.

Typically, such problem involving exponents converge to very few remainders. Here is an explanation.
Look at the nature of 9^2 = 81. And 81^x will always end with unit digit as 1.
So
9^381 = 9*(9^380) = 9*(81^190)
whis will have 9 in units place.
So dividing this number by 5 will leave 4 as remainder.

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Re: Remainder   [#permalink] 30 Oct 2008, 23:30
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What is the remainder when 9^381 is divided by 5? How do you

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