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

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freedom128
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How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)| [#permalink] New post   Updated on: 06 Aug 2019, 21:24
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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
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B

Originally posted by freedom128 on 06 Aug 2019, 10:35.
Last edited by freedom128 on 06 Aug 2019, 21:24, edited 7 times in total.
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Re: How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)| [#permalink] New post   06 Aug 2019, 11:00
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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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Re: How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)| [#permalink] New post   28 Mar 2020, 06:54
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chondro48 wrote:
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


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
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Re: How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)| [#permalink] New post   28 Aug 2020, 22:32
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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

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Re: How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)| [#permalink] New post   29 Aug 2020, 17:38
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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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Re: How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)| [#permalink] New post   04 Nov 2023, 22:05
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧




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

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Re: How many pairs of positive integers (x,y) can satisfy |(x + 3)(y - 3)| [#permalink]
04 Nov 2023, 22:05
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