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virtualanimosity
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Numbers 6 06 Nov 2009, 06:06
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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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kalpeshchopada7
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Numbers 6 06 Nov 2009, 06:19
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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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Numbers 6 06 Nov 2009, 06:33
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kalpeshchopada7
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.
Show more

CLARIFICATION

l is actually mod.....its not a variable...ie value of (x-4) has to be positive
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Numbers 6 06 Nov 2009, 06:46
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virtualanimosity
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
Show more

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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Numbers 6 06 Nov 2009, 06:57
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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.



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