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But there’s something in me that just keeps going on. I think it has something to do with tomorrow, that there is always one, and that everything can change when it comes. http://aimingformba.blogspot.com

Last edited by aiming4mba on 04 Aug 2010, 22:19, edited 1 time in total.

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.

Given: mn=p+1. Question: is m+n=p?

(1) Both m and n are prime numbers. If m=n=2, then p=3 and the answer is NO, as m+n=2+2=4\neq{p=3}but if m=2 and n=3 then p=5 and the answer is YES, as m+n=2+3=p=5. Not sufficient.

(2) p + 1 is even --> p=odd. No info about m and n. Not sufficient.

(1)+(2) Examples from (1) are still valid (as in both examples p=odd), hence we still have two different answers. Not sufficient.

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.

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.

Re: Any other method to solve this test? [#permalink]
02 Feb 2011, 19:39

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(n-1) = (n + 1)?

m = \frac{n+1}{n-1}?

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.

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.