1 + 2m + 2mn + n = 45

AJWILL wrote:If m and n are positive integers and 1 + 2m + 2mn + n = 45, which of the following can be the value(s) of m + n ?
(I) 7
(II) 8
(III) 9

[A] I & II
I & III
[C] II & III
[D] I, II, III
[E] Only II


whats the smart logic?

eagleeye wrote:

AJWILL wrote:If m and n are positive integers and 1 + 2m + 2mn + n = 45, which of the following can be the value(s) of m + n ?
(I) 7
(II) 8
(III) 9

[A] I & II
I & III
[C] II & III
[D] I, II, III
[E] Only II


whats the smart logic?

1 + 2m + 2mn + n = 45
=> 1(1+2m) + n(2m+1) =3*3*5
=> (2m+1)*(n+1) = 3*3*5

Now, since m and n are positive integers, we can match the product on RHS to LHS.

Now factor pairs of 3*3*5 are
3*15
5*9
45*1
We can discard checking 45*1, because it will give large values for m and n, and we have sum of m and n limited to 9 in the options.

Let's check the other 2:

(2m+1)*(n+1) = 3*15
Case 1: n+1 = 15, clearly n=14 is too large for our options. We need not check
Case 2: 2m+1 = 15, n+1 = 3 => m =7, n = 2. Hence m+n = 9. So 9 is one of the solutions.

(2m+1)*(n+1) = 5*9
Case 1: n+1 =9, 2m+1 = 5 => n=8, m=2. sum is 10 which isn't one of our options.
Case 2: 2m+1=9, n+1 = 5, => m=4, n=4. sum is 8 which is one of the solutions.

Hence only 8,9 are among the answers.

C is the correct answer.