Question

A five-digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways this can be done is:

• A. 122
• B. 210
• C. 216
• D. 217

Hint:

A number is divisible by 3 when its digit sum is divisible by 3, so first find which digit to drop from the six given.

Answer 1

The Correct Option is C

Work through the same idea but count the invalid cases first and subtract, a cleaner route for the zero-containing set.

1. Set with no zero, {1,2,3,4,5}: every arrangement of these 5 digits gives a valid five-digit number, since there is no zero to worry about, giving \(5! = 120\).
2. Set with zero, {0,1,2,4,5}: count all arrangements first, \(5! = 120\), then remove the ones where 0 sits in the leading spot. Fixing 0 first and arranging the other 4 digits in the remaining spots gives \(4! = 24\) invalid numbers.
3. Subtract invalid from total: \(120 - 24 = 96\) valid numbers from this set.

Total across both digit sets = \(120 + 96 = 216\), matching option C.

Let's summarize:

• Splitting into a zero-free case and a zero-included case, then handling the leading-digit restriction only where it applies, avoids double work.

Both digit sets together give 216 valid five-digit multiples of 3.