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# M25-35

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Math Expert
Joined: 02 Sep 2009
Posts: 52431

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16 Sep 2014, 00:24
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Difficulty:

35% (medium)

Question Stats:

66% (01:00) correct 34% (01:02) wrong based on 392 sessions

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If $$m$$ and $$n$$ are positive integers, then is $$m$$ an even integer?

(1) $$\frac{m}{n}$$ is an odd integer.

(2) $$m + n$$ is an even integer.

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Math Expert
Joined: 02 Sep 2009
Posts: 52431

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16 Sep 2014, 00:24
Official Solution:

(1) $$\frac{m}{n}$$ is an odd integer. Hence $$m=n*\text{odd}$$. Now, if $$n=\text{odd}$$ then $$m=\text{odd}$$ but if $$n=\text{even}$$ then $$m=\text{even}$$. Not sufficient.

(2) $$m+n$$ is an even integer. Either both are odd or both are even. Not sufficient.

(1)+(2) Still the same two cases are possible: either both are odd (for example $$m=3$$ and $$n=1$$) or both are even (for example $$m=2$$ and $$n=2$$). Not sufficient.

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30 Dec 2014, 08:02
Hey,

I didn't understand your point about [1]. This part in specific: Hence m=n*odd.

[1] says that m/n is odd and an integer. This should mean that m is a multiple of n, otherwise it wouldn't be an integer, right?
So, in order to have an integer as the answer, both m and n should be even or odd, with m>n.
But, since we now that it is odd, then both m and n should be odd, sth like 9/3=3. Which means that both m and n are odd, so we get that m is an odd integer and not an even integer. So, this is sufficient.

Where am I making the mistake..?
Math Expert
Joined: 02 Sep 2009
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31 Dec 2014, 03:52
pacifist85 wrote:
Hey,

I didn't understand your point about [1]. This part in specific: Hence m=n*odd.

[1] says that m/n is odd and an integer. This should mean that m is a multiple of n, otherwise it wouldn't be an integer, right?
So, in order to have an integer as the answer, both m and n should be even or odd, with m>n.
But, since we now that it is odd, then both m and n should be odd, sth like 9/3=3. Which means that both m and n are odd, so we get that m is an odd integer and not an even integer. So, this is sufficient.

Where am I making the mistake..?

This is very simple: $$\frac{m}{n}$$ is an odd integer --> $$\frac{m}{n}=odd$$ --> $$m=n*\text{odd}$$. If $$n=\text{odd}$$ then $$m=\text{odd}$$ (for example m=n=1) but if $$n=\text{even}$$ then $$m=\text{even}$$ (for example m=n=2). Not sufficient.
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Status: Math is psycho-logical
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31 Dec 2014, 04:23
1
Hmmm, I just saw my mistake. For some reason, I didn't think of options such as 10/2=5, 6/2=3, but only such as 4/2=2, 8/2=4. So, because my examples were of even numbers ending in another even number, I rejected the possibility of the integers being even. This is why picking numbers can sometimes be tricky...

Thank you.
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Joined: 02 Jan 2016
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12 May 2018, 22:29
Hi Bunuel,

I applied the generalized approach, Even No.* Even No. or Even No. * Odd No. = Even No.
The same outcomes are also applicable for division too, Because I had read about this rule somewhere, So I assumed, here as the outcome of Division is Odd, both No.s have to be Odd.

Is this rule for division correct ?
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12 May 2018, 23:34
hero_with_1000_faces wrote:
Hi Bunuel,

I applied the generalized approach, Even No.* Even No. or Even No. * Odd No. = Even No.
The same outcomes are also applicable for division too, Because I had read about this rule somewhere, So I assumed, here as the outcome of Division is Odd, both No.s have to be Odd.

Is this rule for division correct ?

Even/Even might be even, odd or not an integer at all. For example, 4/2 = 2 = even, 2/2 = 1 = odd, 2/4 = 1/2 = not an integer.

Even/odd might be even, or not an integer at all. For example, 6/3 = 2 = even, 4/3 = not an integer.

Odd/even is not an integer.

Odd/odd is either an odd integer or not an integer at all. 49/7 = 7 = odd and 7/49 is not an integer.
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16 Jul 2018, 08:20
1
Bunuel wrote:
If $$m$$ and $$n$$ are positive integers, then is $$m$$ an even integer?

(1) $$\frac{m}{n}$$ is an odd integer.

(2) $$m + n$$ is an even integer.

Solution :

(1) m/n odd --> Say m/n = k then K odd.

m = n*k --> m can be odd or even depending on whether n is odd or even.

So (1) not sufficient.

(2) m + n even then both can be even or both can be odd

So (2) not sufficient.

(1) and (2) together

m + n = kn + n = n* (k+1)

k+1 is even

so n can be even or odd. Hence m can be even or odd.

(1) and (2) together not sufficient.

Ans E)
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16 Jul 2018, 19:18
Statement 1: consider (2,6) and (3,9) the answer is yes and no
Statement 2: consider the same sets the answer is yes and no

If 2 statements are combined: the answer can be both yes and no

Then E
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Joined: 25 Jul 2017
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16 Jul 2018, 20:26
Bunuel wrote:
If $$m$$ and $$n$$ are positive integers, then is $$m$$ an even integer?

(1) $$\frac{m}{n}$$ is an odd integer.

(2) $$m + n$$ is an even integer.

For this type of questions, I always prefer solving by examples.

From Statement 1: m/n is odd take m=(1,2) and n = (1/2) respectively.
=> in both scenarios, m/n = 1 but m could be either 1 or 2 = Not sufficient
From Statement 2: It simply states both m & n are either odd or even - not Suffieint

even if we combine both, that is both m & n are odd (take 1,1) or even (take 2,2)- we don't get definite answer.

Hence, Answer is E
Re: M25-35 &nbs [#permalink] 16 Jul 2018, 20:26
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# M25-35

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