How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)|

Question

How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)| = 4 ?

A. 1
B. 2
C. 3
D. 4
E. None

firas92 · Accepted Answer

Since x and y are integers, (x+3) and (y-3) are also integers

Additionally since x is a positive integer, (x+3)≥4 so the minimum value that (x+3) can take is 4 and if (x+3) is greater than 4, we have no integer solution to get |(x+3)*(y-3)|=4

Therefore 4 is the only value that (x+3) can take which means x=1

Now we have 2 ways to get |(x+3)*(y-3)|=4 which are |4*1| and |4*-1|

(y-3) can be 1 or -1 for positive integer values of y (y=2 or y=4)

Therefore our possible solutions are (1,2) and (1,4)

We have 2 positive integer pairs

Answer is (B)

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Kinshook · Answer

Asked: How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)| = 4 ?

4 = 2*2 = 1*4

|x+3| = 1; |y-3| = 4; Since x= -3+-1 ; 0 solutions
|x+3| = 4; |y-3| = 1; (x,y) = {(1,2),(1,4)} ; 2 solutions
|x+3| = 2; |y-3| = 2; Since x=-3+-2 ; 0 solutions

IMO B

CrackverbalGMAT · Answer

Since (x,y) are positive (x,y) > 0 (0 is neither positive or negative)

|(x + 3) (y - 3)| = 4

Case 1:

2 * 2 = 4

Opening the modulus:

-2 = x + 3 = 2, Therefore x = -1 and x = -5.

We can stop here itself, as x has to be positive. This does not give a solution.

Case 2:

4 * 1 = 4

Opening the modulus:

- 4 = x + 3 = 4

Solving, we get x = 1 and x = -7.

x = 1 is a possible option

-1 = y - 3 = 1

Solving, we get y = 4, y = 2.

So we get 2 solutions (x = 1, y = 4) and (x = 1, y = 2)

Case 3:

1 * 4 = 4

Opening the modulus:

-1 = x + 3 = 1

Solving, we get x = -4 and -2. Again, we can stop here itself, as x has to be positive. This does not give a solution.

Therefore there are 2 solutions as listed in Case 2.

Option B

Arun Kumar

paras7424 · Answer

B?

The keyword over here Pair of Positive Integers. Hence x+3 will always b greater than 3 and can attain a value of max 4.

Y-3 can attain a value of either -1 or 1 since we have a modulus outside the brackets. Hence y will be either 2 or 4.

Thua we get 2 pairs 1,2 and 1,4.

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BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

https://www.youtube.com/watch?v=jW0u5xzeq1w

We need to find how many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)| = 4 ?

x and y are positive integers
=> x + 3 will be ≥ 3

x + 3 and y - 3 are two integers whose product is 4 and x + 3 ≥ 3
=> Possible combinations can be |4 * 1| and |4 * -1|
=> x + 3 = 4
=> x = 1

For y we have two cases

-   Case 1:  y - 3 = 1  
=> y = 4

-   Case 2:  y - 3 = -1  
=> y = 2

So, we have two pairs (4,4) and (4,2)

So,  Answer will be B 
Hope it helps!

Absolute Values Theory and Solved Problems   Playlist Link here

Watch the following video to learn the Basics of Absolute Values

https://www.youtube.com/watch?v=huFiBk8PaGY

ravitejab1 · Answer

Important thing to remember is zero is an integer but it is neither a +ve nor a -ve integer.

bumpbot · Answer

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How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)|

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For y we have two cases
-Case 1: y - 3 = 1 => y = 4	-Case 2: y - 3 = -1 => y = 2

GMAT Problem Solving (PS) Questions

How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)|

Prep Toolkit

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