Question

597**6 Is Divisible By Both 3 And 11. The Nonzero Digits In The Hundred’s And Ten’s Places Are Respectively?​

Answer 1

Answer

Let the given number be 597xy6.

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

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

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

=>(27+x+x+3)

=>(30+2x)

=> x = 3 and y = 6.