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16 Dec 2019, 02:43
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Difficulty:
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Question Stats:
60% (01:39) correct
40%
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wrong
based on 242
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If m and k are non-zero integers, is m a multiple of k?
(1) \(\frac{m^2 + m}{k}\) is an integer.
(2) \(m = 2k^2 − 3k\)
Are You Up For the Challenge: 700 Level Questions
(1) \(\frac{m^2 + m}{k}\) is an integer.
(2) \(m = 2k^2 − 3k\)
Are You Up For the Challenge: 700 Level Questions
firas92
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16 Dec 2019, 04:52
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Bunuel
(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)
16 Dec 2019, 05:05
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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.
Posted from my mobile device
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.
Posted from my mobile device
