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The integers m and p are such that 2<m<p, and m is not a

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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink]

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New post 06 Jun 2018, 23:51
chauhanvibhu wrote:
from 1st hint what if i take m=6 and p=10 then remainder comes out to be 4 .
then answer shouldn't be (E) ?

Can someone please explain where i am going wrong ?


YES, the remainder comes out to be 4, But the question is asking whether the remainder is > 1 or not? 4 is > 1, so answer to the question asked is YES. Sufficient to answer still. No matter which example you take, you have to take both even values such that one is not a factor of other, and remainder will thus be either 2 or 4 or 6... always greater than 1.
Re: The integers m and p are such that 2<m<p, and m is not a   [#permalink] 06 Jun 2018, 23:51

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The integers m and p are such that 2<m<p, and m is not a

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