597**6 Is Divisible By Both 3 And 11. The Nonzero Digits In The Hundr

1.	597**6 Is Divisible By Both 3 And 11. The Nonzero Digits In The Hundred’s And Ten’s Places Are Respectively?
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 =&GT; y=x+3 27+x+y) =>(27+x+x+3) =>(30+2x) => x = 3 and y = 6. 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.

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

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

=&GT; y=x+3 27+x+y)

=>(27+x+x+3)

=>(30+2x)

=> x = 3 and y = 6.

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.