Question

$476xy0$ is divisible by both $3$ and $11$. The non-zero digits in the hundred's and ten's places are respectively:

• A. $7$ and $4$
• B. $7$ and $5$
• C. $8$ and $5$
• D. None of these

Answer 1

Answer: (C)

$8$ and $5$

Let the given number be $476$ $xy$ $0$.
Then $(4+7+6+x+y+0)=(17+x+y)$ must be divisible by $3$
And, $(0+x+7)-(y+6+4)=(x=y=3)$ must be either $0$ or $11$.
$x-y-3=0$ $\Rightarrow$ $y=x+3$
$(17+x+y)=(17+x+x-3)=(2x+14)$
$\Rightarrow$ $x=2$ or $x=8$
$\therefore$ $x=8$ and $y=5$