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If m and k are non-zero integers, is m a multiple of k?

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If m and k are non-zero integers, is m a multiple of k? [#permalink] New post 16 Sep 2012, 03:28
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If m and k are non-zero integers, is m a multiple of k?

(1) (m^2+m)/k is an integer.
(2) m=2k^2−3k
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Re: If m and k are non-zero integers, is m a multiple of k [#permalink] New post 16 Sep 2012, 04:44
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navigator123 wrote:
If m and k are non-zero integers, is m a multiple of k?
(1) (m2+m)/k is an integer.
(2) m=2k2−3k


(1) \frac{m^2+m}{k}=\frac{m(m+1)}{k}.
m and m+1 are consecutive integers, so they don't have any common factor except 1 (they are co-prime).
So, k must be a factor of either m or m+1.
Not sufficient.

(2) Dividing through by k gives \frac{m}{k}=2k-3 which is an integer, therefore k must be a divisor of m.
Sufficient.

Answer B.
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Re: If m and k are non-zero integers, is m a multiple of k [#permalink] New post 16 Sep 2012, 09:37
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navigator123 wrote:
If m and k are non-zero integers, is m a multiple of k?
(1) (m2+m)/k is an integer.
(2) m=2k2−3k



The catch is what eva said,
Two consecutive integers are co prime !!

if that doesnt strike you, put in some value.


Remember, the difference between an identity and an equation.
An equation is tru for certain values of variable, whereas an identity is true for all values of variables.
Here what we are given is an Identity on the set of integers so put in a value which proves this wrong

And also remember oone more thing, if a value you put into it satisfies doesn't mean that it is true, but if it doesn't satisfiy it does mean that it is not true


Here put m = 3 and take k = 4
m(m+1) = 3*4 = 12, K divides 12 but k doesnt divide 3, hence not sufficient.
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Re: If m and k are non-zero integers, is m a multiple of k? [#permalink] New post 16 Dec 2013, 22:48
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Re: If m and k are non-zero integers, is m a multiple of k? [#permalink] New post 03 Feb 2014, 05:58
Pretty straight forward B.

Little bit different explanation than above posts.

(1) \frac{(m^2+m)}{k} is integer.

\frac{m^2}{k}+\frac{m}{k} is integer. It is possible that \frac{m^2}{k} and \frac{m}{k} each is integer and their sum is integer. But it is also possible that both of these are non-integers that add up to an integer. Hence not sufficient.

(2) m = 2K^2-3k

m = k(2k-3)

\frac{m}{k} = 2k-3. Clearly an integer. sufficient.
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Re: If m and k are non-zero integers, is m a multiple of k?   [#permalink] 03 Feb 2014, 05:58
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