Latest Elitmus Aptitude Question SOLUTION: Z^2 = 100^2 - (X^2), X from 0 to 100, for how many values of X ,Z is divisible by 6.

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(#M40037748) ELITMUS Numerical Ability question Query Keep an EYE

Z^2 = 100^2 - (X^2),
X from 0 to 100,
for how many values of X ,Z is divisible by 6.

Asked In Elitmus arun pandey (9 years ago)

Unsolved Read Solution (24)

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Read Solution (24) :

Saraswathy  (9 years ago)

Ans ll be 0...
z^2 = 100^2 - x^2
z = sqrt (100^2 - x^2)
z = sqrt(100+x)(100-x)
z = sqrt(100+x) * sqrt(100-x)
So check for all the Perfect square values which is >100 nd (100+x)=121 -> (100-x)=79 -> 11*sqrt(79) not divisible by 6
x=44 -> (100+44)=144 -> (100-x)=56 -> 12*sqrt(56) not divisible by 6
Like that check for all perfect square values... None ll be divisible by 6...
So ans is 0

Ans : 0

Is this solution Helpfull? Yes (45) | No (17)  

REYANSH  (9 years ago)

ans is 3
for x=28 z=96, x=80 z=60 , x=100 z=0

i don't know the procedure to solve the question. i had made program for this in c++ adn got the ans from there.

Is this solution Helpfull? Yes (35) | No (6)  

Ajay kumar Jangir  (9 years ago)

x values are 16 only...
because Z always divided by 6 means Z always contains of 6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96 so you can insert values of X between 0 to 100 only multiple of 6 but max 96 after that u get values are negative

Is this solution Helpfull? Yes (6) | No (14)  

aditi bose  (9 years ago)

3 values of x

Is this solution Helpfull? Yes (5) | No (3)  

Kumar Siddhartha  (9 years ago)

take different values of z as multiples of 6
6^2=100^2-x^2
x^2=100^2-6^2
x^2=(100-6)(100+6)
if the ans will be a perfect square then the ans will be the value of x for z to be divisible by 6
following these steps we will have to check up to 16 factors of 6 i.e. 96 will be the last one and one of the right ans
you will finally get the ans as 3 for the number of values of x for which z is divisible by 6

Is this solution Helpfull? Yes (5) | No (4)  

Mithun Kumar Kesari  (9 years ago)

z^2=100^2-x^2
X^2=100^2-Z^2

since, z is divisible by 6, it is a multiple of 6 also.

so the values of Z^2 =0, 6^2, 12^2, 18^2, 24^2, ...........96^2
for all these values, check for which value of Z^2 . X^2 is a perfect squares..

after check we only get z=0, 60, 96... three values only.

Is this solution Helpfull? Yes (5) | No (1)  

ssdv prasad  (9 years ago)

ans is 0
for z=sqrt(100+x)*sqrt(100-x)
x value 96 is possible but it gives z value 28 which is not divisible by 6.

Is this solution Helpfull? Yes (3) | No (2)  

Prakhar Gupta  (9 years ago)

From 0 to 100.

At x = 0 then z = 100
x = 28 then z = 96
x = 60 then z = 80
x = 80 then z = 60
x = 28 then z = 96
x = 60 then z = 80
x = 80 then z = 60
x = 96 then z = 28
x = 100 then z = 0

Only 4 values of x are there in which Z is divisible by 6
i.e.
x = 28 then z = 96
x = 60 then z = 80
x = 80 then z = 60
x = 100 then z = 0

But According to my Program Logic

for(int x = 0; x

Is this solution Helpfull? Yes (3) | No (7)  

Ehtesham  (9 years ago)

x can be 28 ,80;
first take x=28.
z^2=100^2 - 28^2
=10000 - 784
=9216
so, z=sqrt(9216) =96 which is divisible by 6;

now
take x=80.
z^2=100^2 - 80^2
=10000 - 6400
=3600
so, z=sqrt(3600) =60 which is divisible by 6;

so there will be two values for which Z is divisible by 6;

Is this solution Helpfull? Yes (3) | No   

kirti sinha  (9 years ago)

x=8
z^2=36
z=6
1 value

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

nilanjan das  (9 years ago)

z^2=(100+x)(100-x).as z shd be devisable by 6 means z62 must be devisable by 36.now the minumum value of (100-x) =0 and max of(100+x)=200.so we have to find how many numbers within 0 to 200 are devisable by 36.so the answer is 6

Is this solution Helpfull? Yes (2) | No   

LOKESH SAMANT  (9 years ago)

0

check for these conditions
z^2=(100-x)*(100+x)
and u can check for the condition z^2+x^2=100^2

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

ATUL TRIPATHI  (9 years ago)

z^2=100^2-x^2
z^2=(100+x)(100-x)

now Z^2 will be divisible by 6 if either of the 2 brackets will be divisible
and range of 100+x will be upto 200
range of 100-x will be upto 100
so there will be 198/6=33 values of x which will cause z^2 to be divisible by 6.

Is this solution Helpfull? Yes (1) | No   

saurabh agarwal  (9 years ago)

Ans wil be 5...
z^2 = 100^2 - x^2
z = sqrt (100^2 - x^2)
z = sqrt(100+x)(100-x)
z = sqrt(100+x) * sqrt(100-x)
if(100+x)or(100-x)is multiple of 36 then it must be divide by 6
when x=64,28,8,44,80,then it must be multiple of 36.

Ans : 5

Is this solution Helpfull? Yes (1) | No (3)  

GAURAV KUMAR SINGH  (7 years ago)

x can be 28 ,80;
first take x=28.
z^2=100^2 - 28^2
=10000 - 784
=9216
so, z=sqrt(9216) =96 which is divisible by 6;
now
take x=80.
z^2=100^2 - 80^2
=10000 - 6400
=3600
so, z=sqrt(3600) =60 which is divisible by 6;

Is this solution Helpfull? Yes (1) | No   

Shikha Sinha  (9 years ago)

ONE VALUE 10

Is this solution Helpfull? Yes | No (4)  

Sahil  (9 years ago)

Ans. is 1 value ie. 0

for value of x=0 only z is divisible by 6.

Is this solution Helpfull? Yes | No (3)  

Prakhar Gupta  (9 years ago)

3 Values of X if x>=0 && x

Is this solution Helpfull? Yes | No (1)  

Prakash Singh  (9 years ago)

ans will be 16

Is this solution Helpfull? Yes | No (1)  

shaifali singh  (9 years ago)

what is the right solution guys ??plz explain also..

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Jaskaran Singh  (8 years ago)

Equation should be
Z =100^2-x^2

Not z^2

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Rahul Ranjan  (8 years ago)

EHTESHAM U R DAMN RIGHT...THANKS

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Santanu das  (8 years ago)

ans wil be 3.
z^2 can be 16,36,96 whch are divisible by 6.

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rohit  (7 years ago)

ans is 66 values
33 for +ve nd 33 for -ve

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