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

Key Concepts You Need

1. Divisibility by 3: A number is divisible by 3 if the sum of its digits is a multiple of 3.
2. Permutations without repetition: If you arrange n distinct items in a line, there are $n!$ ways. If some positions are restricted (like the first digit cannot be 0), we subtract the invalid cases.
3. Choosing which digit to leave out: Because we must use 5 of the 6 given digits, we first decide which single digit is excluded, then arrange the remaining five.

Logical Chain of Thought

1. Compute the total digit sum of all six: $15$. This sum is divisible by 3.
2. Leaving out one digit: The new 5-digit sum is $15 - d$ where $d$ is the omitted digit. For $15 - d$ to remain divisible by 3, $d$ itself must be divisible by 3 (because 15 already is).
3. Identify possible omissions: Among 0,1,2,3,4,5 only $0$ and $3$ are divisible by 3.
4. Case-wise permutation counting
   • Case 1: Omit 0 – Remaining digits $\{1,2,3,4,5\}$. No zero problem. Ways $= 5! = 120$.
   • Case 2: Omit 3 – Remaining digits $\{0,1,2,4,5\}$. Total permutations $=5! =120$ but subtract the $4! = 24$ cases where 0 is first: $120-24=96$.
5. Add both cases: $120+96 = 216$.

Thus, there are 216 valid 5-digit numbers.

Imagine you have 6 different coloured blocks numbered 0,1,2,3,4,5 and you want to build a 5–block long train.

1. Train must have 5 blocks, so one colour will be left out.
2. The engine (first block) cannot be 0, because then it is not a real 5-digit number (it would look like a 4-digit train).
3. Divisible by 3 rule: A number is divisible by 3 if the sum of its digits is a multiple of 3.
4. The sum of all six blocks is 0+1+2+3+4+5 = 15, which already goes evenly into 3.
5. If we leave out one block, the new sum must still go evenly into 3. That only happens if the block we leave out is itself a multiple of 3 – so we can skip 0 or 3.
6. Now just count how many ways we can arrange the remaining 5 blocks each time:
   • Skip 0 ⇒ use 1,2,3,4,5 → everything allowed: 5! = 120 ways.
   • Skip 3 ⇒ use 0,1,2,4,5 → but 0 can’t be first. Total 5! = 120 minus the 24 where 0 is first ⇒ 96 ways.
7. Add them: 120 + 96 = 216 different trains (numbers) you can make.

So, 216 ways in total!

Step-by-Step Solution

1. Total sum of all six digits
   $0 + 1 + 2 + 3 + 4 + 5 = 15$ Since 15 is divisible by 3, the sum of the chosen five digits will be $15 - d$ where $d$ is the digit we leave out.

2. Find permissible digit to omit
   $15 - d \text{ divisible by } 3 \Longrightarrow d \text{ itself divisible by } 3$
   Possible $d$: $0,\;3$

3. Case 1: Omit 0

   Remaining digits: $1,2,3,4,5$
   Number of 5-digit permutations: $5! = 120$

4. Case 2: Omit 3

   Remaining digits: $0,1,2,4,5$
   Total permutations (if no restriction): $5! = 120$
   Invalid permutations with 0 in the first place: fix 0, arrange other 4 digits ⇒ $4! = 24$
   Valid permutations: $120 - 24 = 96$

5. Total required numbers

   $120 + 96 = 216$

Answer: 216