**Question:**

597**6 is divisible by both 3 and 11. The non-zero digits in the Hundred's and ten's places are respectively:

597366....

597366

597366

Then (5+9+7+x+y+6)=(27+x+y) must be divisible by 3

And, (6+x+9)-(y+7+5)=(x-y+3) must be either 0 or divisible by 11.x-y+3=0

=> y=x+3 27+x+y)

=>(27+x+x+3)

=>(30+2x)

=> x = 3 and y = 6.

597366, 597696

597366, 59769

597366, __597696__

3 & 6

Munna is __rgt__

Sorry my ans is __wrong__