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If x and y are positive integers and 1 + x + y +xy = 21

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If x and y are positive integers and 1 + x + y +xy = 21 [#permalink]

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If x and y are positive integers and 1 + x + y +xy = 21, what is the value of x?

(1) y>3
(2) y=6

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Re: If x and y are positive integers and 1 + x + y +xy = 21 [#permalink]

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Good question. +1.

If x and y are positive integers and 1 + x + y +xy = 21, what is the value of x?

\(1 + x + y +xy = 21\) --> \((x+1)(y+1)=21\) --> either \(x+1=3\) and \(y+1=7\) OR \(x+1=7\) and \(y+1=3\). Notice that \(x+1=1\) and \(y+1=21\) OR \(x+1=21\) and \(y+1=1\) is not possible since in this case either \(x\) or \(y\) equals zero and we are told that \(x\) and \(y\) are positive integers.

Now, from \(x+1=3\) and \(y+1=7\) --> \(x=2\) and \(y=6\) AND from \(x+1=7\) and \(y+1=3\) --> \(x=6\) and \(y=2\).

(1) y>3 --> \(y=6\), so \(x=2\). Sufficient.
(2) y=6 --> \(x=2\). Sufficient.

Answer: D.

Hope it's clear.
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Re: If x and y are positive integers and 1 + x + y +xy = 21 [#permalink]

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We can factor 1 + x + y + xy = 21 to determine that (1 + x)(1 + y) = 21. The product of the integers (1 + x) and (1 + y) is 21, which has the factors 1, 3, 7, and 21. The factor pair 1 and 21 is disqualified because neither (1 + x) nor (1 + y) could equal 1, as that would make one of the positive integers x or y equal to zero. We therefore determine that (1 + x) could equal 3 or 7, and conversely, (1 + y) could equal 7 or 3 such that their product is 21. Knowing that x will equal either 2 or 6, we can rephrase the question: “Is the value of x equal to 2 or 6?”

(1) SUFFICIENT: If y > 3, then it must be true that y = 6 and x = 2.

(2) SUFFICIENT: If y = 6, then x = 2.

The correct answer is D.
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Re: If x and y are positive integers and 1 + x + y +xy = 21 [#permalink]

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Hi,

We have, 1 + x + y +xy = 21 (where x & y are natural numbers/positive integers)
or x(1+y) = 20 -y
or x = (20-y)/(1+y)

Using (1)
y > 3, i.e., y = 4, 5, 6...
when y=4, x = (20-4)/5 = 16/5 (not a natural number)
when y=5, x = (20-5)/6 = 15/6 (not a natural number)
when y=6, x = (20-6)/7 = 14/7 = 2
when y=7, x = (20-7)/8 = 13/8 (not a natural number), and for remaining values of y, there would be no positive integral value of x.
so, (x,y) = (2,6). Sufficient

Using (2)
y = 6, => x = (20-6)/(1+6) = 2
so, (x,y) = (2,6). Sufficient.

Thus, Answer is (D)

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Re: If x and y are positive integers and 1 + x + y +xy = 21 [#permalink]

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Re: If x and y are positive integers and 1 + x + y +xy = 21 [#permalink]

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New post 04 Mar 2017, 06:06
Prompt analysis
x an y are the integers such that
1 +x +y +xy =21 or (1+x)(1+y) = 21
therefore (1+x)(1+y) could be
1 x 21. hence x =0, y = 20 (1)
3 x 7. hence x =2, y = 6 (2)
7 x 3. hence x =6, y = 2 (3)
21 x 1. . hence x =20, y = 0 (4)

Translation
To find the exact solution among the 4 options we need
1# the range of x and y
2# exact value of x and y

Statement analysis
St 1: x>3. hence option (3) and (4) applicable. INSUFFICIENT
St 2: y =6. Hence option (2) only is applicable. ANSWER

Option B
Re: If x and y are positive integers and 1 + x + y +xy = 21   [#permalink] 04 Mar 2017, 06:06
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