5-digit Number Puzzle

For $a$ 474 $b$ to be divisible by 63 it must be also divisible by 9. When an integer is divisible by 9, the sum of its digits is also divisible by 9. Since $4+7+4 = 15$ , $a+b =$ 3 or 12; and the possible solutions are:

$\implies \begin{cases} a+b=3 & \implies \begin{cases} a=1, b=2 & \color{#3D99F6} 63 \ | \ 14742 \small \text{ a solution} \\ a=2, b=1 & \color{#D61F06} 63 \nmid 24741 \small \text{ not a solution} \end{cases} \\ a+b = 12 & \implies \begin{cases} a=3, b=9 & \color{#D61F06} 63 \nmid 34749 \small \text{ not a solution} \\ a=4, b=8 & \color{#D61F06} 63 \nmid 44748 \small \text{ not a solution} \\ a=5, b=7 & \color{#3D99F6} 63 \ | \ 54747 \small \text{ a solution} \\ a=6, b=6 & \color{#D61F06} 63 \nmid 64746 \small \text{ not a solution} \\ a=7, b=5 & \color{#D61F06} 63 \nmid 74745 \small \text{ not a solution} \\ a=8, b=4 & \color{#D61F06} 63 \nmid 84744 \small \text{ not a solution} \\ a=9, b=3 & \color{#D61F06} 63 \nmid 94743 \small \text{ not a solution} \end{cases} \end{cases}$

We note that the only acceptable option is $\boxed{3}$ .