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

Step 1: Recall the divisibility rule for 3.
A number is divisible by 3 only when the sum of its digits is divisible by 3. We must pick 5 digits out of the 6 given digits (0, 1, 2, 3, 4, 5) and arrange them.

Step 2: Find the sum of all six digits.
\(0+1+2+3+4+5 = 15\), which is already divisible by 3. So dropping one digit that is itself divisible by 3 keeps the remaining sum divisible by 3.

Step 3: Identify which digit to drop.
Digits divisible by 3 in the list are 0 and 3. So there are two possible 5-digit sets: \(\{1,2,3,4,5\}\) (drop 0) and \(\{0,1,2,4,5\}\) (drop 3).

Step 4: Count arrangements for the set without 0.
\(\{1,2,3,4,5\}\) has no zero, so all \(5! = 120\) arrangements are valid five-digit numbers.

Step 5: Count arrangements for the set with 0.
\(\{0,1,2,4,5\}\) has 0 in it, and a five-digit number cannot start with 0. Total arrangements \(5! = 120\), arrangements with 0 first \(4! = 24\), so valid arrangements = \(120 - 24 = 96\).

Step 6: Add both cases. \[ 120 + 96 = 216 \]

Final Answer:
There are 216 such five-digit numbers. \[ \boxed{216} \]