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If p and q are positive integers, what is the remainder when 92p × 5p

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If p and q are positive integers, what is the remainder when 92p × 5p [#permalink]

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New post 30 Nov 2016, 21:12
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If p and q are positive integers, what is the remainder when 9^2p × 5^(p+q) + 11^q × 6^pq is divided by 10?

A)0
B)1
C)3
D)4
E)5
[Reveal] Spoiler: OA

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Re: If p and q are positive integers, what is the remainder when 92p × 5p [#permalink]

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New post 30 Nov 2016, 21:52
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Here is the Solution =>
let n = 9^2p × 5^(p+q) + 11^q × 6^pq
Rule used => The remainder whenever an integer is divided by 10 is same as its units digit.
As a side note => In order to get the remainder with 2,5 or 10 --> We just need the units digit.
So we actually need the units digit of n
Applying the concept of cyclicity

Units digit of n=> 1*5+1*6=>5+6 =>1
Hence the units digit will be one and so will the remainder when n is divided by 10.
Hence B

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Re: If p and q are positive integers, what is the remainder when 92p × 5p [#permalink]

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New post 14 Dec 2017, 06:57
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Re: If p and q are positive integers, what is the remainder when 92p × 5p   [#permalink] 14 Dec 2017, 06:57
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