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805+ (Hard)|   Algebra|      
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ArathiKrishnan
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If x and y are two distinct integers such that xy + x + y + 1 = 20 Updated on: 30 Jan 2022, 09:04
Originally posted by ArathiKrishnan on 30 Jan 2022, 06:01.
Last edited by Bunuel on 30 Jan 2022, 09:04, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Difficulty:

95% (hard)

Question Stats:

37% (02:00) correct 63% (02:13) wrong based on 60 sessions

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If x and y are two distinct integers such that xy + x + y + 1 = 20, what is the value of x?

(1) x = 2n, where n is an integer
(2) x > y
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E
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If x and y are two distinct integers such that xy + x + y + 1 = 20 Updated on: 30 Jan 2022, 19:19
Originally posted by Kinshook on 30 Jan 2022, 06:46.
Last edited by Kinshook on 30 Jan 2022, 19:19, edited 2 times in total.
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Asked: If x and y are two distinct integers such that xy + x + y + 1 = 20, what is the value of x?
xy + x + y + 1 = (x+1)(y+1) = 20 = 2^2*5
(x,y) = {(3,4),(4,3),(1,9),(9,1),(0,19),(19,0),(-5,-6),(-6,-5),(-3,-11),(-11,-3),(-2,-21),(-21,-2)}

1) x =2n, where N is an integer
If x is even, then the possible solutions are when (x,y) = {(4,3),(0,19),(-6,-5),(-2,-21)}
NOT SUFFICIENT

2) x>y
(x,y) = {(4,3),(9,1),(19,0),(-5,-6),(-3,-11),(-2,-21)}
NOT SUFFICIENT

(1) + (2)
1) x =2n, where N is an integer
2) x>y
(x,y) = {(4,3),(-2,-21)}
NOT SUFFICIENT

IMO E
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If x and y are two distinct integers such that xy + x + y + 1 = 20 30 Jan 2022, 06:50
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ArathiKrishnan
If x and y are two distinct integers such that xy + x + y + 1 = 20, what is the value of x?

1) x =2n, where N is an integer
2) x>y

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xy + x + y + 1 = 20
(x+1)(y+1) = 20
So (x+1) and (y+1) are factors of 20


From 1: x = 2n, ie even; so (x+1) is odd
(x+1) is an odd factor of 20
=> x + 1 = 1 or 5 => x = 0 or 4 - not unique


From 2: x > y:
x+1 = 20 and y+1 = 1 ie x = 19 and y = 0
OR
x+1 = 10 and y+1 = 2 ie x = 9 and y = 1; etc.
Thus, x is not unique

Thus, either statement is not sufficient

Combining:
x + 1 = 1, y + 1 = 20 => x = 0 and y = 19 - doesn't satisfy that x > y; so rejected
OR
x + 1 = 5, y + 1 = 4 => x = 4 and y = 3 - satisfies x > y; so accepted
Thus, the solution is unique

Answer C
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If x and y are two distinct integers such that xy + x + y + 1 = 20 30 Jan 2022, 09:05
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Kinshook
Asked: If x and y are two distinct integers such that xy + x + y + 1 = 20, what is the value of x?
xy + x + y + 1 = (x+1)(y+1) = 20 = 2^2*5
(x,y) = {(3,4),(4,3),(1,9),(9,1),(0,19),(19,0)}

1) x =2n, where N is an integer
If x is even, then only possible solution is when x = 4 and y=3
SUFFICIENT

2) x>y
(x,y) = {(4,3),(9,1),(19,0)}
NOT SUFFICIENT

IMO A
Show more

sujoykrdatta
ArathiKrishnan
If x and y are two distinct integers such that xy + x + y + 1 = 20, what is the value of x?

1) x =2n, where N is an integer
2) x>y

Posted from my mobile device

xy + x + y + 1 = 20
(x+1)(y+1) = 20
So (x+1) and (y+1) are factors of 20


From 1: x = 2n, ie even; so (x+1) is odd
(x+1) is an odd factor of 20
=> x + 1 = 1 or 5 => x = 0 or 4 - not unique


From 2: x > y:
x+1 = 20 and y+1 = 1 ie x = 19 and y = 0
OR
x+1 = 10 and y+1 = 2 ie x = 9 and y = 1; etc.
Thus, x is not unique

Thus, either statement is not sufficient

Combining:
x + 1 = 1, y + 1 = 20 => x = 0 and y = 19 - doesn't satisfy that x > y; so rejected
OR
x + 1 = 5, y + 1 = 4 => x = 4 and y = 3 - satisfies x > y; so accepted
Thus, the solution is unique

Answer C
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I think both of you are missing negative solutions for the equation. No?
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If x and y are two distinct integers such that xy + x + y + 1 = 20 30 Jan 2022, 19:20
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Bunuel

Please check the revised solution. You are correct. I missed negative solutions in the first attempt.

Bunuel
Kinshook
Asked: If x and y are two distinct integers such that xy + x + y + 1 = 20, what is the value of x?
xy + x + y + 1 = (x+1)(y+1) = 20 = 2^2*5
(x,y) = {(3,4),(4,3),(1,9),(9,1),(0,19),(19,0)}

1) x =2n, where N is an integer
If x is even, then only possible solution is when x = 4 and y=3
SUFFICIENT

2) x>y
(x,y) = {(4,3),(9,1),(19,0)}
NOT SUFFICIENT

IMO A

sujoykrdatta
ArathiKrishnan
If x and y are two distinct integers such that xy + x + y + 1 = 20, what is the value of x?

1) x =2n, where N is an integer
2) x>y

Posted from my mobile device

xy + x + y + 1 = 20
(x+1)(y+1) = 20
So (x+1) and (y+1) are factors of 20


From 1: x = 2n, ie even; so (x+1) is odd
(x+1) is an odd factor of 20
=> x + 1 = 1 or 5 => x = 0 or 4 - not unique


From 2: x > y:
x+1 = 20 and y+1 = 1 ie x = 19 and y = 0
OR
x+1 = 10 and y+1 = 2 ie x = 9 and y = 1; etc.
Thus, x is not unique

Thus, either statement is not sufficient

Combining:
x + 1 = 1, y + 1 = 20 => x = 0 and y = 19 - doesn't satisfy that x > y; so rejected
OR
x + 1 = 5, y + 1 = 4 => x = 4 and y = 3 - satisfies x > y; so accepted
Thus, the solution is unique

Answer C

I think both of you are missing negative solutions for the equation. No?
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If x and y are two distinct integers such that xy + x + y + 1 = 20 10 Apr 2024, 04:33
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