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M25-35

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M25-35  [#permalink]

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New post 16 Sep 2014, 01:24
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A
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D
E

Difficulty:

  35% (medium)

Question Stats:

65% (01:00) correct 35% (01:02) wrong based on 382 sessions

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Re: M25-35  [#permalink]

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New post 16 Sep 2014, 01:24
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.


Answer: E
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Re: M25-35  [#permalink]

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New post 30 Dec 2014, 09:02
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..?
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Re: M25-35  [#permalink]

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New post 31 Dec 2014, 04:52
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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Re: M25-35  [#permalink]

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New post 31 Dec 2014, 05:23
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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Re: M25-35  [#permalink]

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New post 12 May 2018, 23:29
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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Re: M25-35  [#permalink]

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New post 13 May 2018, 00:34
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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Re: M25-35  [#permalink]

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New post 16 Jul 2018, 09:20
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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Re: M25-35  [#permalink]

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New post 16 Jul 2018, 20:18
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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Re: M25-35  [#permalink]

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New post 16 Jul 2018, 21:26
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
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Re: M25-35 &nbs [#permalink] 16 Jul 2018, 21:26
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