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1 + 2m + 2mn + n = 45

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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1 + 2m + 2mn + n = 45

by AJWILL » Sun Jul 22, 2012 3:10 am
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?
Last edited by AJWILL on Sun Jul 22, 2012 11:15 am, edited 1 time in total.
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by eagleeye » Sun Jul 22, 2012 8:35 am
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.
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by AJWILL » Sun Jul 22, 2012 10:29 am
i wish y brain could function like yours.
thanks though
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by Jim@StratusPrep » Mon Jul 23, 2012 6:46 am
beagleeye beat me to it.

However, this illustrates a key point with the GMAT... always think about factoring!
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by truplayer256 » Sun Aug 05, 2012 2:02 pm
Factoring might ease the problem up for us:
1 + 2m + 2mn + n = 45
1 + 2m + n(2m + 1) = 45
(1 + 2m)(1 + n) = 45
Let's list the possible combinations of both the factors:
1 + 2m = 9 or 1 + n = 5 or vice versa: m = 4, n = 4 or n = 8, m = 2
Similarly:
1 + 2m = 15 or 1 + n = 3 or vice versa: m = 7, n = 2 or n = 14, m = 1

Note that the only possible values of m+n can be 8 or 9.

Choose C.
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by mjones » Mon Aug 06, 2012 1:53 pm
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.


How did you get from "1(1+2m) + n(2m+1)" to "(2m+1)*(n+1)" exactly? That's the only part of this entire question I'm still not clear on.
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