w 2 y 2 x 2 z 2 where w x y z all are odd number this equation is

even*even is always divisible by 4
for eg 2*28
42*38
these are always divisible by 4
10 years agoHelpfull: Yes(32) No(4)

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..
10 years agoHelpfull: Yes(19) No(19)
d. 2
odd+odd=even
even*even = even
10 years agoHelpfull: Yes(10) No(11)
c) 4

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

=> even*even/4
10 years agoHelpfull: Yes(3) No(1)
ans is 2
odd*odd=odd
odd+odd=odd or even
even*even=even
10 years agoHelpfull: Yes(1) No(1)
no need to see question it is clear from options that 2 is the answer.
10 years agoHelpfull: Yes(1) No(1)
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
10 years agoHelpfull: Yes(0) No(3)
c. always divisible by 4
10 years agoHelpfull: Yes(0) No(0)
why 4 kratika ? and why not 2
10 years agoHelpfull: Yes(0) No(0)

(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