597**6 is divisible by both 3 and 11. The non-zero digits in the Hundred’s and ten’s places are respectively: Choices:
A) 3 and 6	B) 7 and 9	C) 2 and 6	D) 4 and 7	E) None
Solution : 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)=(x-y+3) must be either 0 or divisible by 11. x-y+3=0   =&gt; y=x+3 27+x+y)   =&gt;(27+x+x+3)   =&gt;(30+2x)   =&gt; x = 3 and y = 6.   3 and 6nswer 3 and 6
<result>A</result>
597**6 is divisible by both 3 and 11. The non-zero digits in the Hundred’s and ten’s places are respectively: