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If m and n are positive integers and mn = p + 1, is m + n = [#permalink]
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04 Aug 2010, 22:49
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If m and n are positive integers and mn = p + 1, is m + n = p ? a. Both m and n are prime numbers. b. p + 1 is even.
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Last edited by aiming4mba on 04 Aug 2010, 23:19, edited 1 time in total.



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Re: DS question [#permalink]
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04 Aug 2010, 23:35
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Re: DS question [#permalink]
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03 Sep 2010, 05:32
sikalvag wrote: Hi, I dont know whether my approach is correct or wrong but I suspect anser is C. From St1: i know they are prime numbers and from St2: i get P as odd number  mn = p + 1 from here, in order to p+1 to be even out of m and n one should be 2. so i get m= (p+1)/2 (if n=2). => m+n =p => (p+1)/2 + 2 = p => defenitly not equal to P.
Please let me know if my approach was wrong.
Thanks OA is given in the first post, under the spoiler and it's E. In my post above there are 2 cases given satisfying the stem and both statements and giving different answers to the question, thus proving that answer is E: If \(m=n=2\), then \(p=3=odd\) and the answer is NO, as \(m+n=2+2=4\neq{p=3}\); If \(m=2\) and \(n=3\) then \(p=5=odd\) and the answer is YES, as \(m+n=2+3=p=5\). Also why "(p+1)/2 + 2 = p => defenitely not equal to P" (the red part)? If you solve it for \(p\) you'll get \(p=5\) so \(n=2\) and \(m=3\). Hope it helps.
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Re: DS question [#permalink]
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04 Aug 2010, 22:58
I think it's C. what's the OA ?



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Re: DS question [#permalink]
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03 Sep 2010, 04:48
Hi, I dont know whether my approach is correct or wrong but I suspect anser is C. From St1: i know they are prime numbers and from St2: i get P as odd number  mn = p + 1 from here, in order to p+1 to be even out of m and n one should be 2. so i get m= (p+1)/2 (if n=2). => m+n =p => (p+1)/2 + 2 = p => defenitly not equal to P.
Please let me know if my approach was wrong.
Thanks



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Re: DS question [#permalink]
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21 Sep 2010, 06:49
Bunuel, thanks again for a concise approach.



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Any other method to solve this test? [#permalink]
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02 Feb 2011, 19:24
If m and n are positive integers and mn = p + 1, is m + n = p ? S1: Both m and n are prime numbers. S2: p + 1 and m are both even A. S1 sf B S2 sf C both A and B together sf D. Each sf E. Neither sf nor together sf
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Re: Any other method to solve this test? [#permalink]
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02 Feb 2011, 19:39
Test 1: M(2), N(3) P(5)
Meets criteria
Test 2: M(2), N(7) P(13)
Fails criteria
All criteria fail to meet test 2
Therefore (E)



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Re: Any other method to solve this test? [#permalink]
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02 Feb 2011, 19:59
i am not fan of the method where we have to plug in numbers unless we have to. Any other suggestions?
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Re: Any other method to solve this test? [#permalink]
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02 Feb 2011, 20:39
Let us analyze what the question is asking prior to looking at the statements given. We know that:
\(mn = p + 1\)
We are asked does:
\(m + n = p?\)
Using what we know, we can rearrange this question as follows:
\(m + n = p?\)
\(m + n = mn  1?\)
\(mn  m = n + 1?\)
\(m(n1) = (n + 1)?\)
\(m = \frac{n+1}{n1}?\)
Since we know that m and n are both positive integers, n can not be greater than 3, otherwise m will result in a value between 1 and 2. We also n can not be 1. Therefore, this leaves two distinct possibilities:
\((m,n) = (2,3),(3,2)\)
Now let's move on to solving the question knowing these conditions.
Statement 1: Both m and n are prime numbers.
2 and 3 are both prime numbers, but so are 11 and 17. We need to know specifically that m and n are 2 and 3.
Therefore, not sufficient.
Statement 2: p + 1 and m are both even.
All this really tells us is that m is even. Given the initial condition that mn = p + 1, if either m or n are given to be even, it follows that p + 1 must be even as well. Hence, the distinct subset of (2,3) still exists, as well as various other possibilities of an even number and any other number.
Therefore, not sufficient.
Both Statements Together
We know that m and n are prime numbers, and that m is even. So m must be 2. Unfortunately, n is only defined to be a prime number. This could be 3 (in which case the statement is satisfied), but it could be any other prime number as well.
Therefore, not sufficient.
Answer: E



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Re: DS question [#permalink]
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03 Feb 2011, 04:25
It took me between 3 and 4 minutes to think and answer this: Solved it using numbers eventually. If m and n are positive integers and mn = p + 1, Q: m + n = p ? 1. Both m and n are prime numbers. 2. p + 1 is even. mn = p + 1 So, p is one less than mn 1. Started with lowest prime numbers m=2, n=2 > mn = 4, p=3: m + n = 4; 4<>3. Ans: No m=2, n=3 > mn= 6, p=5: m + n = 5; 5=5. Ans: Yes Not sufficient. 2. p + 1 is even p is odd. Used the same data set and disapproved: m=2, n=2 > mn = 4, p=3(odd): m + n = 4; 4<>3. Ans: No m=2, n=3 > mn= 6, p=5(odd): m + n = 5; 5=5. Ans: Yes Not sufficient. Together: Same data set. Not sufficient. Ans: E
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Re: DS question [#permalink]
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05 Feb 2011, 04:48



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Re: If m and n are positive integers and mn = p + 1, is m + n = [#permalink]
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05 Jan 2014, 13:40
aiming4mba wrote: If m and n are positive integers and mn = p + 1, is m + n = p ? a. Both m and n are prime numbers. b. p + 1 is even. This boils down to Is m+n = mn1 Statement 1 m,n are prime numbers Let's number pick. Mind you. if m and n are 2 and 3 then yes If m and 3 are 2 and 5 then no Insuff Statement 2 p+1 is even, then p is odd We get is mn even? Both together mn could be even as well as odd depending on whether the number 2 is included as one of both Hence answer is E Cheers! J



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Re: If m and n are positive integers and mn = p + 1, is m + n = [#permalink]
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