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If x and y are positive integers and 1 + x + y + xy = 15, wh

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If x and y are positive integers and 1 + x + y + xy = 15,  [#permalink]

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New post 12 Jun 2008, 03:45
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If x and y are positive integers and 1 + x + y + xy = 15, what is the value of x + y?

A. 3
B. 5
C. 6
D. 8
E. 9

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Re: Arithmatic Problem  [#permalink]

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New post 12 Sep 2010, 08:11
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udaymathapati wrote:
What's the approach to solve following example.
If x and y are positive integers and 1 + x + y + xy = 15, what is the value of x + y?
A. 3
B. 5
C. 6
D. 8
E. 9


\(1+x+y+xy=15\) --> \((x+1)(y+1)=15=3*5\);

As \(x\) and \(y\) are positive integers then \(x+1=3\) and \(y+1=5\), or vise-versa (x+1=1 and y+=15 or vise-versa is not possible as in this case one of the unknowns becomes zero and we are told that both unknowns are positive).

So \(x+1=3\) and \(y+1=5\), or vise-versa --> no need to solve for \(x\) and \(y\), just add them --> \(x+1+y+1=3+5\) --> \(x+y=6\).

Answer: C.
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Re: If x and y are positive integers and 1 + x + y + xy = 15,  [#permalink]

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New post 13 Jun 2008, 02:23
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mrwaxy wrote:
If x and y are positive integers and 1 + x + y + xy = 15, what is the value of x + y?
A. 3
B. 5
C. 6
D. 8
E. 9


1+x+y+xy=15

1+x + y(x+1)=15

(1+x)(1+y)=15

(1+x)(1+y)=3*5

this implies x=2 and y = 4

therefore x+y = 2 + 4 = 6

hence C is the answer
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Re: If x and y are positive integers and 1 + x + y + xy = 15,  [#permalink]

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New post 12 Jun 2008, 06:49
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mrwaxy wrote:
If x and y are positive integers and 1 + x + y + xy = 15, what is the value of x + y?
A. 3
B. 5
C. 6
D. 8
E. 9


Try each answer choices.
For A: 1 + 3 + xy = 15; xy = 11 (impossible, 11 prime number. 1 + 11 doesn't equal 3)
For B: 1 + 5 + xy = 15; xy = 9 (no combination of xy = 9 and x + y = 5)
For C: 1 + 6 + xy = 15; xy = 8 (x + y = 6; x = 2, y = 4 or x = 4, y = 2)
For D: 1 + 8 + xy = 15; xy = 6 (no combination of xy = 6 and x + y = 8)
For E: 1 + 9 + xy = 15; xy = 5 (impossible, 5 prime number. 1 + 5 doesn't equal 9)

Therefore, answer C.
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Re: If x and y are positive integers and 1 + x + y + xy = 15,  [#permalink]

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New post 13 Jun 2008, 00:25
2
mrwaxy wrote:
If x and y are positive integers and 1 + x + y + xy = 15, what is the value of x + y?
A. 3
B. 5
C. 6
D. 8
E. 9



1 + x + y + xy = 15

1 + x + y (1+ x) = 15

(1 + x) (1+ y) = 3*5

x+y=(3-1)+(5-1)
=2+4=6

C. 6
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Re: Arithmatic Problem  [#permalink]

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New post 12 Sep 2010, 06:58
I think this is how it is calculated;

Factors of 15 are 1,3,5,and 15.

So 1 + x + y +xy = (1+x)(1+y) = 15. Basically find product of two numbers that is equal to 15. This can be factorized as follows

1 + x = 3 and 1 + y = 5 (as product of 3x5 = 15)
-> x = 2 and y = 4

OR

1 + x = 15 and 1 + y = 1 (as product of 15x1 = 15)
-> x=14 and y=0

However problem indicates that x,y>0, so value of x = 2 and y = 4.

Value of x+y = 2+4 = 6

Ans C
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Re: Arithmatic Problem  [#permalink]

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New post 12 Sep 2010, 08:26
I went by answe choices.
x+y+xy=14 (given)
If x+y = 3 then x=1, y 2 = cannot satisfy 14.
If x+y=5, x=4, y=1 cannot satisfy 14. x=2 and y=3 ..ruled out
x+y=6 = I got the answer

I liked the equation approach as its clean. Although I could solve it in less than a minute
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Re: If x and y are positive integers and 1 + x + y + xy = 15, wh  [#permalink]

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New post 23 Jul 2015, 22:31
mrwaxy wrote:
If x and y are positive integers and 1 + x + y + xy = 15, what is the value of x + y?

A. 3
B. 5
C. 6
D. 8
E. 9


Easiest way to solve this question seems substitution because the values can't be too big as the multiplication of those two Positive Integers x and y must be less than 15 as per the given expression 1 + x + y + xy = 15

Values of x and y have to be less than 3 and 5 as 3*5=15

Let's try with 2 and 3
1+2+3+2*3 = 12

i.e. One of the numbers must be a little bigger

Let's try with 2 and 4
1+2+4+2*4 = 15 BINGO!!!

i.e. x+y = 2+4 = 6

Answer: option C
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Re: If x and y are positive integers and 1 + x + y + xy = 15, wh  [#permalink]

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New post 08 Nov 2017, 16:41
mrwaxy wrote:
If x and y are positive integers and 1 + x + y + xy = 15, what is the value of x + y?

A. 3
B. 5
C. 6
D. 8
E. 9


Note that the expression y + xy can be factored as y(1 + x). Let’s simplify the given equation:

1 + x + y(1 + x) = 15

(1 + x) + y(1 + x) = 15

The two terms on the left side of the equation have (1 + x) in common, so we can factor it out:

(1 + x)(1 + y) = 15

Since x and y are positive integers, so are (1 + x) and (1 + y). We should investigate the different ways of writing 15 as a product of positive integers. Note that 15 can be expressed as a product of positive integers in two ways: 1 x 15 (scenario one) or 3 x 5 (scenario two).

For scenario one, we have either 1 + x = 1 and 1 + y = 15 OR 1 + x = 15 and 1 + y = 1. Note that in either case, one of x or y equals 0; since x and y are given as positive, this scenario is ruled out.

For scenario two, we have either 1 + x = 3 and 1 + y = 5, which means x = 2 and y = 4 OR 1 + x = 5 and 1 + y = 3, meaning x = 4 and y = 2. In either case, x + y = 6.

Answer: C
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Re: If x and y are positive integers and 1 + x + y + xy = 15, wh  [#permalink]

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New post 24 Dec 2017, 14:53
Can someone show me the details of this step?

1+x + y(x+1)=15

(1+x)(1+y)=15
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Re: If x and y are positive integers and 1 + x + y + xy = 15, wh  [#permalink]

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New post 24 Dec 2017, 16:55
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Quote:
Can someone show me the details of this step?

1+x + y(x+1)=15

(1+x)(1+y)=15

xlgoh1992 , sure.

1) 1+x + y(x + 1) = 15

Factor out (1 + x) from the first two terms (because 1 + x is identical to x + 1 by which y is multiplied)

2) 1 (1 + x) + y(x+1)
Notice that if you multiplied 1(1 + x) you would be back to: 1 + x

3) Note that (1 + x) has already been factored out in this term: y(x + 1)

1 (1 + x) + y (x + 1)

4) Combine what factoring out (1 + x) left: 1 and + y
(1 + x) (1 + y) = 15

Hope that helps.
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Re: If x and y are positive integers and 1 + x + y + xy = 15, wh &nbs [#permalink] 24 Dec 2017, 16:55
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