If m and n are positive integers and mn = p + 1, is m + n = p ?

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.

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

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)

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.

Therefore, not sufficient.

Answer: E

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

This boils down to

Is m+n = mn-1

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

If m and n are positive integers and mn = p + 1, is m + n = p ?
a. Both m and n are prime numbers.
m =2, n = 3; m = 3, n = 5...only (2)(3) = 5 + 1 suffices. (3)(5) = 14 +1; (2)(5) = 9 +1. Thus except, case 1 - but rest don't add up. So NOT SUFF.
b. p + 1 is even. No clue about m,n. Not Suff.

St 1 and 2: p +1 = even; m and n = both prime. Again only one case (2)(3) = 5 +1 works; rest don't - (2)(5) = 9 +1; (2)(13)= 26 (25 +1). So not sufficient. Ans E.

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