Question

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

Answer 1

Correct Answer: 63

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.