Forming Five-Digit Numbers Divisible by 3 Without Repetition

The problem asks us to find the number of five-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, and 5 without repeating any digit, such that the number is divisible by 3.

Understanding Divisibility by 3

A number is divisible by 3 if the sum of its digits is divisible by 3.

Identifying the Set of Digits

We have a set of 6 digits: {0, 1, 2, 3, 4, 5}. We need to choose exactly 5 of these digits to form a five-digit number.

The sum of all 6 available digits is: $$0 + 1 + 2 + 3 + 4 + 5 = 15$$

Since 15 is divisible by 3, if we remove a digit that is a multiple of 3 (or leaves a remainder of 0 when divided by 3), the sum of the remaining 5 digits will still be divisible by 3. If we remove a digit that is not a multiple of 3, the sum of the remaining 5 digits will not be divisible by 3.

We need the sum of the five chosen digits to be divisible by 3. Let S be the sum of the 6 digits (S=15). Let d be the digit we remove. The sum of the remaining 5 digits will be S - d = 15 - d. We need 15 - d to be divisible by 3.

Since 15 is divisible by 3, 15 - d is divisible by 3 if and only if d is divisible by 3.

The digits available are {0, 1, 2, 3, 4, 5}. The digits divisible by 3 in this set are 0 and 3.

Therefore, to form a five-digit number divisible by 3, we must exclude either the digit 0 or the digit 3 from the original set of 6 digits.

Case 1: Exclude the digit 0

The set of 5 digits used is {1, 2, 3, 4, 5}. The sum of these digits is $1+2+3+4+5 = 15$, which is divisible by 3.

We need to form a five-digit number using these 5 distinct digits. Since none of the digits is 0, any arrangement of these 5 digits will form a valid five-digit number.

The number of ways to arrange 5 distinct digits is the number of permutations of 5 items taken 5 at a time, denoted as $P(5, 5)$ or $5!$.

Number of ways in Case 1 = $$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$

Case 2: Exclude the digit 3

The set of 5 digits used is {0, 1, 2, 4, 5}. The sum of these digits is $0+1+2+4+5 = 12$, which is divisible by 3.

We need to form a five-digit number using these 5 distinct digits. However, the first digit of a five-digit number cannot be 0.

We can calculate the number of arrangements in two ways:

Method 1: Total permutations minus permutations starting with 0

• Total permutations of the 5 digits {0, 1, 2, 4, 5} is $5! = 120$.
• Permutations that start with 0: If 0 is fixed in the first position, the remaining 4 digits {1, 2, 4, 5} can be arranged in the remaining 4 positions in $4!$ ways. $$4! = 4 \times 3 \times 2 \times 1 = 24$$
• The number of five-digit numbers (that do not start with 0) is the total number of permutations minus the number of permutations starting with 0.
• Number of ways in Case 2 = $5! - 4! = 120 - 24 = 96$.

Method 2: Filling positions considering restrictions

• For the first digit (ten thousands place), we can choose any digit except 0. From the set {0, 1, 2, 4, 5}, the available choices for the first digit are {1, 2, 4, 5}. There are 4 options.
• For the second digit, we can choose any of the remaining 4 digits (including 0). There are 4 options.
• For the third digit, we can choose any of the remaining 3 digits. There are 3 options.
• For the fourth digit, there are 2 remaining options.
• For the fifth digit, there is 1 remaining option.
• Number of ways in Case 2 = $4 \times 4 \times 3 \times 2 \times 1 = 4 \times 4! = 4 \times 24 = 96$.

Both methods give the same result: 96 ways for Case 2.

Total Number of Ways

The total number of five-digit numbers divisible by 3 is the sum of the ways from Case 1 and Case 2.

Total ways = Ways in Case 1 + Ways in Case 2

Total ways = $120 + 96 = 216$

Therefore, there are 216 ways to form a five-digit number divisible by 3 using the digits 0, 1, 2, 3, 4, and 5 without repetition.

Revision Table: Number Formation and Divisibility by 3

Let's recap the key concepts used in solving this problem.

Additional Information: Counting Principles

This problem uses fundamental counting principles from combinatorics.

• The Sum Rule: If an event can occur in $m$ ways and another event can occur in $n$ ways, and the two events cannot occur simultaneously, then the number of ways that either event can occur is $m+n$. In our problem, forming a number from the set {1, 2, 3, 4, 5} and forming a number from {0, 1, 2, 4, 5} are mutually exclusive events (they use different sets of digits), so we add the number of ways from each case.
• The Product Rule: If a procedure can be broken down into a sequence of tasks, and there are $n_1$ ways to perform the first task, $n_2$ ways to perform the second task after the first has been completed, and so on, then there are $n_1 \times n_2 \times \dots$ ways to perform the procedure. This is implicitly used in calculating permutations ($n! = n \times (n-1) \times \dots \times 1$) and in the alternative method for Case 2 ($4 \times 4 \times 3 \times 2 \times 1$).

Understanding divisibility rules and basic counting principles like permutations and the sum/product rules is crucial for solving problems involving number formation and arrangements.