Question

The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is ________.

Answer 1

For odd number, unit place will be $1, 3, 5, 7$ or $9$.
So, $xy1, xy3, xy5, xy7, xy9$ are the type of numbers.
If $xy1$ then:
$x + y = 6, 13, 20 …$ Cases are required
i.e., $6 + 6 + 0 + … = 12$ ways
If $xy3$ then:
$x + y = 4, 11, 18, ….$ Cases are required
i.e., $4 + 8 + 1 + 0 … = 13$ ways
Similarly for $xy5$, we have
$x + y = 2, 9, 16, …$
i.e., $2 + 9 + 3 = 14$ ways
for $xy7$ we have
$x + y = 0, 7, 14, ….$
i.e., $0 + 7 + 5 = 12$ ways
And for $xy9$ we have
$x + y = 5, 12, 19 …$
i.e., $5 + 7 + 0 … = 12$ ways

So, total $63$ ways.