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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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New post 11 Jun 2012, 07:20
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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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New post 11 Jun 2012, 07:40
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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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New post 11 Jun 2012, 07:29
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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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New post 03 Mar 2017, 13:23
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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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New post 04 Mar 2017, 05:06
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
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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 18 Mar 2018, 21:59
BDSunDevil wrote:
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

[Searched, but couldn't find]


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

1 + x + y +xy = 21
⇔ (x+1)(y+1) = 21
⇔ x+1 = 3, y+1 = 7 or x+1 = 7, y+1 = 3 since x and y are positive integers.
⇔ x=2, y=6 or x=6, y=2

Condition 1)
y > 3 ⇒ y = 6 ⇒ x = 2
The condition 1) is sufficient.

Condition 2)
y = 6 ⇒ x = 2
The condition 2) is sufficient.

Therefore, D is the answer.
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If x and y are positive integers and 1 + x + y +xy = 21  [#permalink]

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New post Updated on: 09 Sep 2018, 00:34
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 x or y equal to zero (and they must be positive integers). Therefore, (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, 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)

Originally posted by alphak3nny001 on 08 Sep 2018, 20:35.
Last edited by Bunuel on 09 Sep 2018, 00:34, edited 2 times in total.
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If x and y are positive  [#permalink]

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New post 08 Sep 2018, 21:01
alphak3nny001 wrote:
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


Given, \((x+1)(y+1) = 21 = 3*7\)

Statement I:

For \((y+1)(x+1) = 21\) to satisfy \(y = 6, x = 2\) (As x, y are positive Integer).

Statement II:

\(y = 6, x = 2\)

Hence, D.
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If x and y are positive  [#permalink]

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New post 08 Sep 2018, 21:54
alphak3nny001 wrote:
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


Two ways...

1) Algebraic..
\(1+x+y+xy=(1+x)+y(1+x)=(1+y)(1+x)=21\)
Now since X and y are positive integers so X+1 and y+1 will be greater than 1..
Thus 21=1*21 not possible
Only other possibility is 21=3*7..
Thus (1+x)(1+y)=3*7...
So x and y are 3-1=2 and 7-1=6

We have to find which value is which variable..

1) y>3
Therefore y cannot be 2, X is 2
Sufficient

2) y=6
Thus X is 2
Sufficient

D

2) Trial

1+x+y+xy=21.....x+y+xy=20
1) y>3
So least value of y is 4, substitute
x+4+4x=20.....5x=16 but then x is not an integer
Let y=5.....5+x+5x=20....6x=15...NO
Let y =6.....6+x+6x=20....7x=14...X=2... possible
As we increase y X decrease, so lowest possible of X is 1..
So 1+y+y=20....2y=19...not possible
Thus X=2
Sufficient
2) y=6
1+6+x+6x=21.......X=2
Sufficient

D
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If x and y are positive   [#permalink] 08 Sep 2018, 21:54
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If x and y are positive integers and 1 + x + y +xy = 21

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