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![How many sets of positive integer solutions does the equation [m]2(x –](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "How many sets of positive integer solutions does the equation [m]2(x –"){kind=icon}
14 Jan 2022, 03:15
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C
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Difficulty:
75%
(hard)
Question Stats:
58% (02:12) correct
42%
(02:30)
wrong
based on 215
sessions
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How many sets of positive integer solutions does the equation \(2(x – 4) + y + z = 0\) have?
(A) 14
(B) 12
(C) 9
(D) 8
(E) 4
Are You Up For the Challenge: 700 Level Questions
(A) 14
(B) 12
(C) 9
(D) 8
(E) 4
Are You Up For the Challenge: 700 Level Questions
14 Jan 2022, 08:23
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Bunuel
Breaking Down the Info:
Since we have to plug in positive integers for all numbers, we must create negative terms in order to balance the equation.
Then \(x - 4\) must be negative in order to allow y and z to be positive. Thus x = 1, 2 or , 3 only.
x = 1: \(y + z = 6\). y can go from 1 to 5, so that's 5 solutions.
x = 2: \(y + z = 4\). We can choose y from 1 to 3, so that's 3 solutions.
x = 3: \(y + z = 2\). We have only one solution.
Then in total, we have 9 solutions.
Answer: C
20 Mar 2025, 11:30
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I used an alternative method to:
making the equation as 2x+y+z=8
Since the solutions are positive hence neither of the three variables could be zero making 'x' to take only 3 values 1,2&3.
therefore the sets of solution could be as undermentioned :
x=1, y=3, z=3
x=1, y=5, z=1
x=1, y=1, z=5
x=1, y=2, z=4
x=1, y=4, z=2
x=2, y=2, z=2
x=2, y=3, z=1
x=2, y=1, z=3
x=3, y=1, z=1
making the equation as 2x+y+z=8
Since the solutions are positive hence neither of the three variables could be zero making 'x' to take only 3 values 1,2&3.
therefore the sets of solution could be as undermentioned :
x=1, y=3, z=3
x=1, y=5, z=1
x=1, y=1, z=5
x=1, y=2, z=4
x=1, y=4, z=2
x=2, y=2, z=2
x=2, y=3, z=1
x=2, y=1, z=3
x=3, y=1, z=1
