Latest Elitmus Aptitude Question SOLUTION: (w^2+y^2)*(x^2+z^2) where w,x,y,z all are odd number, this equation is always divisible by? a) 20 b) 8 c) 4 d) 2

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(#M40022762) ELITMUS question Quant Keep an EYE

(w^2+y^2)*(x^2+z^2) where w,x,y,z all are odd number, this equation is always divisible by?

a) 20
b) 8
c) 4
d) 2

Asked In Elitmus Narendra Pargain (10 years ago)

Unsolved Read Solution (9)

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kratika jain  (10 years ago)

even*even is always divisible by 4
for eg 2*28
42*38
these are always divisible by 4

Is this solution Helpfull? Yes (32) | No (4)  

pawann raj chauhan  (10 years ago)

so the is solution is as follow... if w is odd no ,and y,x,z is also odd no..and we know that square of odd is always is always odd..(odd^2 + odd^2)*(odd^2+odd^2) and after we know that if odd is add to odd then we get always even..so (odd+odd)*(odd+odd) and after that (even)*(even)...when even is multiply to even it get always even..ex (3^2+5^2)*(7^2+9^2)=(9+25)*(49+81)=34*130=1820 which is even...but we find that all the option above is even..so how we choose a suitable even no..so we have too choose smaller even no..because it can devide all other even no...thats why we choose 2 so the answer is 2. 100% sure..

Is this solution Helpfull? Yes (19) | No (19)  

rituraj rathore  (10 years ago)

d. 2
odd+odd=even
even*even = even

Is this solution Helpfull? Yes (10) | No (11)  

Anurag Kumar Singh  (10 years ago)

c) 4

coz... w^2+y^2=even & x^2+z^2=even ...

=> even*even/4

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

sourav  (10 years ago)

ans is 2
odd*odd=odd
odd+odd=odd or even
even*even=even

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

naga sarath chand  (10 years ago)

no need to see question it is clear from options that 2 is the answer.

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

Bhawana Agrawal  (10 years ago)

d) 2
square of odd is always odd and sum of two odd no is even and even*even is always even. so it will always be divisible by 2

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Vijay Singh Adhikari  (10 years ago)

c. always divisible by 4

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Priyanka Singh  (10 years ago)

why 4 kratika ? and why not 2

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