The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is ________.
Correct Answer: 63
Solution and Explanation
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.
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Concepts Used:
Permutations
A permutation is an arrangement of multiple objects in a particular order taken a few or all at a time. The formula for permutation is as follows:
\(^nP_r = \frac{n!}{(n-r)!}\)
nPr = permutation
n = total number of objects
r = number of objects selected
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