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Numbers 6

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Manager
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Numbers 6 [#permalink] New post 06 Nov 2009, 05:06
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A
B
C
D
E

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(N/A)

Question Stats:

0% (00:00) correct 100% (02:37) wrong based on 1 sessions
How to approach, do we have to individually put in all da values and check, coz there is a possiblity to leaving out a combination

Q.If lx-4l + ly-4l =4, then how many integer values can the set (x,y) have?
a.infinite
b.5
c.16
d.25
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Re: Numbers 6 [#permalink] New post 06 Nov 2009, 05:46
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virtualanimosity wrote:
How to approach, do we have to individually put in all da values and check, coz there is a possiblity to leaving out a combination

Q.If lx-4l + ly-4l =4, then how many integer values can the set (x,y) have?
a.infinite
b.5
c.16
d.25


There could be FOUR cases:

1. x<4, y<4
|x-4|+|y-4|=4 --> -x+4-y+4=4 --> x+y=4. Integer values of (x,y)=(1,3),(2,2),(3,1)
3 sets.

2. x<4, y>4
|x-4|+|y-4|=4 --> -x+4+y-4=4 --> y-x=4. Integer values of (x,y)=(3,7),(2,6),(1,5)
3 sets.

3. x>4, y<4
|x-4|+|y-4|=4 --> x-4-y+4=4 --> x-y=4. Integer values of (x,y)=(5,1),(6,2),(7,3)
3 sets.

4. x>4, y>4
|x-4|+|y-4|=4 --> x-4+y-4=4 --> x+y=12. Integer values of (x,y)=(5,7),(6,6),(7,5)
3 sets.

PLUS we should not forget that x and y can be 4 (but not together):
x=4 --> |x-4|+|y-4|=4 --> |y-4|=4, y=8 or 0. Integer values of (x,y)=(4,8),(4,0)
2 sets.

y=4 --> |x-4|+|y-4|=4 --> |x-4|=4, x=8 or 0. Integer values of (x,y)=(8,4),(0,4)
2 sets.

3+3+3+3+2+2=16

Answer: C.
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Re: Numbers 6 [#permalink] New post 06 Nov 2009, 05:19
I honestly dont feel any need to key in any values. There are 3 variables and the answer asks about values of (x,y) to be in integers. If it is silent about third variable 'l', then the number of combination depends on any integer or FRACTION value of l which will make it impossible to come-up with definite number of x and y into integers.

I would guess it ans A.
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Re: Numbers 6 [#permalink] New post 06 Nov 2009, 05:33
kalpeshchopada7 wrote:
I honestly dont feel any need to key in any values. There are 3 variables and the answer asks about values of (x,y) to be in integers. If it is silent about third variable 'l', then the number of combination depends on any integer or FRACTION value of l which will make it impossible to come-up with definite number of x and y into integers.

I would guess it ans A.


CLARIFICATION

l is actually mod.....its not a variable...ie value of (x-4) has to be positive
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Kudos [?]: 438 [0], given: 5

Re: Numbers 6 [#permalink] New post 06 Nov 2009, 05:57
Ohhh... fine then.

if |x-4| + |y-4|=0, then both x-4 and y-4 have to be positive. which means, there are following possibilities:
if - x-4 and y-4
if - 0 then 4, hence x = 4 and y = 0 or 8
if - 1 then 3, hence x = 3 or 5 and y = 1 or 7
if - 2 then 2, hence x = 2 or 6 and y = 2 or 6
if - 3 then 1, hence x = 1 or 7 and y = 3 or 5
if - 4 then 0, hence x = 0 or 8 and y = 4.

Dont know if there can be more possibilities within positive numbers. Ans should be 16.

OA - C.
Re: Numbers 6   [#permalink] 06 Nov 2009, 05:57
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