Latest Elitmus Aptitude Question SOLUTION: if mod(x-4)+mod(y-4)=4, then how many integer values can the set (x,y) have?

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(#M40038580) ELITMUS Category question Keep an EYE

if mod(x-4)+mod(y-4)=4, then how many integer values can the set (x,y) have?

Asked In Elitmus nidhi gupta (9 years ago)

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RAKESH  (9 years ago)

only 16 integral soln.

|x-4|+|y-4|= 4
x & y can have values from 0 to 8 only.

for x=0 & x =8 , only 1 soln in each case

for x = 1 to 7 , two solns for each values of x.

total = 2 + 2*7 = 16 soln.

check:
if x = 0 , y = 4
if x = 1 , y = 3,5
if x = 2 , y = 2,6
if x = 3 , y = 1,7
if x = 4 , y = 0,8
if x = 5 , y = 1,7
if x = 6 , y = 2,6
if x = 7 , y = 3,5
if x = 8 , y = 0

Is this solution Helpfull? Yes (15) | No (3)  

dayanand giri  (9 years ago)

|x-4|+|y-4|=4
x=0,y=4 (1)
x=1,y=3,5 (2)
x=2,y=2,6 (2)
x=3,y=1,7 (2)
x=4,y=0,8 (2)
x=5,y=1,7 (2)
x=6,y=2,6 (2)
x=7,y=3,5 (2)
x=8,y=4 (1)
than total solution should be 16.

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Arindam Sural  (9 years ago)

Answer will be 14
If we solve the equation then we will get the equation: x+y=12,Now the possible pairs are (0+12),(1+11),(2+10),(3+9),(4+8),(5+7),(6+6),(6+6),(7+5),(8+4),(9+3),(10+2),(11+1),(12+0),which is equal to 14.......pls crct me if m wrng

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Vivek singh bhatnagar  (9 years ago)

16 integral solution


if x = 0 , y = 4
if x = 1 , y = 3,5
if x = 2 , y = 2,6
if x = 3 , y = 1,7
if x = 4 , y = 0,8
if x = 5 , y = 1,7
if x = 6 , y = 2,6
if x = 7 , y = 3,5

-->> if x = 8 , y = 4 ......

Is this solution Helpfull? Yes (2) | No (1)  

Kumar Vikash  (9 years ago)

here is two diffrent variable...then every value of x,y has a unique value...that means it has infinite set of solution.

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Pranoy Bej  (9 years ago)

ANSWER==> 16
there are 16 solutions,
possible combinations are (4+0),(3+1),(2+2), (0+4)
now, for 4+0
X can be 8 or 0 then Y can be 4. So 2 set of values
similarly for 2+2 there are 4 set of values
now adding all the sets of values for all 4 possible combination
we get 16 set of values

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Ranjana Verma  (9 years ago)

ans-40
since there will be two equations. one will be in +ve and other one will be in -ve. and since it is an integer it will take both =ve and -ve set of values.
|x-4|+|y-4|=4 and -|x-4|-|y-4|.

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Neha Agrawal  (9 years ago)

x-4+y-4=4
x+y=12
possible pairs of (x,y) are (0,12),(1,11),(2,10),(3,9),(4,8),(5,7),(6,6),(7,5),(8,4),(9,3),(10,2),(11,1),(12,0)
so there are 13 solutions

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Ranjana Verma  (9 years ago)

ans-38, there will be two equations

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vikash kashyap  (9 years ago)

|x-4|+|y-4|=4
case 1
if x,y>=4
(x-4)+(y-4)=4, x+y=12
At that case possible pair of (x,y) is (4,8),(5,7),(6,6),(7,5),(8,4)

case 2
if x=4
-(x-4)+(y-4)=4, y-x=4
At that case possible pair of (x,y) is (0,4),(1,5),(2,6),(3,)

case 3
if x>=4,y

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