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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]

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New post 16 Dec 2019, 01:43
1
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

64% (01:43) correct 36% (01:51) wrong based on 55 sessions

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If m and k are non-zero integers, is m a multiple of k?  [#permalink]

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New post 16 Dec 2019, 03:52
1
Bunuel wrote:
If m and k are non-zero integers, is m a multiple of k?


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

(2) \(m = 2k^2 − 3k\)


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

\(\frac{m(m+1)}{k}\) is an integer

So either \(m\) is a multiple of \(k\) or \(m+1\) is a multiple of \(k\)

Not sufficient

(2) \(m = 2k^2 − 3k\)

\(m=k(2k-3)\)

Therefore, \(m\) is a multiple of \(k\)

Sufficient

Answer is (B)
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Re: If m and k are non-zero integers, is m a multiple of k?  [#permalink]

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New post 16 Dec 2019, 04:05
I will try to explain the way I understand and solved it. Please pardon me if it's not clear to all.

Statement 1 is clearly insufficient.

If we try to open the expression, we achieve something like

m^2 + m = Kn (where n is an integer value)

m (m+1)= kn.
The values of the expression could be

m= kn, or m+1= kn, so it's practically hard to tell which of them is the multiple. it could be either m or m+1

We can confirm this by plugging in real numbers to check the initial expression.

Let us use M= 4 for our example
4^2 +4 =20.

Our K can be either 2, 4, 5 , 10 or 20 because only these numbers can divide 20 without leaving any remainder.

If K= 5 or 10, then m= 4 is not a multiple of K.

But if k= 4 or 20, then m= 4 becomes a multiple.

So statement 1 clearly is insufficient.

STATEMENT 2: M= 2k^2 - 3k

We can factorise the expression to obtain something like

M = k (2K-3)

We can then subsequently replace the (2k-3) with

Therefore m= kn.

To further confirm, you could plug in real numbers and seethat the expression is clearly correct .

Statement 2 is sufficient

Answer is option B.

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If m and k are non-zero integers, is m a multiple of k?  [#permalink]

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New post 16 Dec 2019, 10:35
Rephrasing the question stem: Is m divisible by k?

statement 1) (m^2+ m ) / k
----> m (m+1) is divisible by K
----> either m or (m+1) or both m &(m+1) is/are divisible by k
----> not sufficient

statement 2) m = 2k^2 - 3k
----> m = k(2k-3)
----> since k is an integer ---> 2k-3 is an integer ----> m = k . Integer
----> sufficient


Answer = B
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If m and k are non-zero integers, is m a multiple of k?   [#permalink] 16 Dec 2019, 10:35
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