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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, 22:51
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.
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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 10 Sep 2018, 05:57
VeritasKarishma would love to see your solution to this problem. Thanks.
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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 11 Sep 2018, 21:47
LM wrote:
The integers m and p are such that 2<m<p, and m is not a factor of p. If r is the remainder when p is divided by m, is r>1?

(1) The greatest common factor of m and p is 2.

(2) The least common multiple of m and p is 30.


m is not a factor of p.

So when p is divided by m, it will not leave remainder 0. We need to find whether the remainder "can be 1" or it "must be greater than 1".

(1) The greatest common factor of m and p is 2.

m and p are both even but p is not a multiple of m. So p will be at least 2 away from m or its multiple. e.g. if m is 4, p cannot be 8 but it can be 10, p cannot be 12 but it can be 14.
So when you divide p by m, you will get at least 2 as remainder. So remainder will be more than 1.
Sufficient.

(2) The least common multiple of m and p is 30.
30 = 2 * 3 *5
m and p could be 5 and 6 (remainder 1)
or m and p could be 6 and 10 (remainder 4)
Not sufficient.

Answer (A)
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Re: The integers m and p are such that 2<m<p, and m is not a &nbs [#permalink] 11 Sep 2018, 21:47

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