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Manager  Status: Trying.... & desperate for success.
Joined: 17 May 2012
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If m and k are non-zero integers, is m a multiple of k?  [#permalink]

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Question Stats: 60% (01:45) correct 40% (01:50) wrong based on 230 sessions

### HideShow timer Statistics 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
Director  Joined: 22 Mar 2011
Posts: 598
WE: Science (Education)
Re: If m and k are non-zero integers, is m a multiple of k  [#permalink]

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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.

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Manager  Joined: 05 Jul 2012
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Re: If m and k are non-zero integers, is m a multiple of k  [#permalink]

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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.
Manager  Joined: 25 Oct 2013
Posts: 143
Re: If m and k are non-zero integers, is m a multiple of k?  [#permalink]

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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]

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_________________ Re: If m and k are non-zero integers, is m a multiple of k?   [#permalink] 14 Jul 2018, 12:30
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