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Re: M2535
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16 Sep 2014, 00:24



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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..?



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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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Re: M2535
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31 Dec 2014, 04:23
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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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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Re: M2535
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16 Jul 2018, 08:20
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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Re: M2535
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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










