Key Concepts and Formulas

• Divisibility Rule for 2: A number is divisible by 2 if its last digit is even.
• Divisibility Rule for 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
• Permutations: The number of ways to arrange $n$ objects is $n!$. If some objects are identical, the number of distinct arrangements is reduced. For $n$ objects with $n_1$ of one kind, $n_2$ of another kind, ..., $n_k$ of the $k$-th kind, the number of distinct arrangements is $\frac{n!}{n_1!n_2!...n_k!}$.

Step-by-Step Solution

Step 1: Applying Divisibility by 2

• What we are doing: Determine the possible values for the units digit.
• Why: A number is divisible by 6 if it's divisible by both 2 and 3. Divisibility by 2 requires the last digit to be even.
• Explanation: The digits we can use are 4, 5, and 9. Only 4 is even. Therefore, the units digit must be 4.

Step 2: Applying Divisibility by 3

• What we are doing: Determine the possible combinations of digits that make the sum of all six digits divisible by 3, knowing that the last digit is 4.

• Why: The divisibility rule for 3 states that the sum of the digits must be divisible by 3.

• Explanation: We have five digits to choose from the set {4, 5, 9}, and the last digit must be 4. Let $n_4$, $n_5$, and $n_9$ be the number of times the digits 4, 5, and 9 appear in the first five places, respectively. We have $n_4 + n_5 + n_9 = 5$. The sum of all six digits is $4n_4 + 5n_5 + 9n_9 + 4$, and this must be divisible by 3. Since $9n_9$ is always divisible by 3, we need $4n_4 + 5n_5 + 4 \equiv 0 \pmod{3}$. Reducing modulo 3, we get $n_4 + 2n_5 + 1 \equiv 0 \pmod{3}$, or $n_4 + 2n_5 \equiv -1 \equiv 2 \pmod{3}$. Also, $n_4 + n_5 \le 5$.

• Now we list the possible non-negative integer solutions to $n_4 + n_5 + n_9 = 5$ and $n_4 + 2n_5 \equiv 2 \pmod{3}$:

  • Case 1: $n_5 = 1$. Then $n_4 \equiv 0 \pmod{3}$, so $n_4 = 0, 3$.

    • If $n_4 = 0$, then $n_9 = 4$. The number of arrangements is $\frac{5!}{0!1!4!} = \frac{120}{24} = 5$.
    • If $n_4 = 3$, then $n_9 = 1$. The number of arrangements is $\frac{5!}{3!1!1!} = \frac{120}{6} = 20$.

  • Case 2: $n_5 = 4$. Then $n_4 \equiv 2 - 8 \equiv 2 - 2 \equiv 0 \pmod{3}$, so $n_4 = 0$.

    • If $n_4 = 0$, then $n_9 = 1$. The number of arrangements is $\frac{5!}{0!4!1!} = \frac{120}{24} = 5$.

  • Case 3: $n_5 = 0$. Then $n_4 \equiv 2 \pmod{3}$, so $n_4 = 2, 5$.

    • If $n_4 = 2$, then $n_9 = 3$. The number of arrangements is $\frac{5!}{2!0!3!} = \frac{120}{12} = 10$.
    • If $n_4 = 5$, then $n_9 = 0$. The number of arrangements is $\frac{5!}{5!0!0!} = 1$.

  • Case 4: $n_5 = 2$. Then $n_4 \equiv 2-4 \equiv -2 \equiv 1 \pmod{3}$, so $n_4 = 1, 4$.

    • If $n_4 = 1$, then $n_9 = 2$. The number of arrangements is $\frac{5!}{1!2!2!} = \frac{120}{4 \cdot 2} = 30$.
    • If $n_4 = 4$, then $n_9 = -1$ (impossible).

  • Case 5: $n_5 = 3$. Then $n_4 \equiv 2-6 \equiv -4 \equiv 2 \pmod{3}$, so $n_4 = 2$.

    • If $n_4 = 2$, then $n_9 = 0$. The number of arrangements is $\frac{5!}{2!3!0!} = \frac{120}{12} = 10$.

• The total number of such arrangements is $5 + 20 + 5 + 10 + 1 + 30 + 10 = 81$.

Step 3: Consider the last digit

• What we are doing: Account for the fact that the last digit is always 4.

• Why: We've calculated the arrangements of the first 5 digits, and the last digit is fixed as 4.

• Explanation: Each of the 81 arrangements of the first 5 digits can be followed by the digit 4 to form a six-digit number divisible by 6. Thus, there are 81 such numbers.

Step 4: Re-evaluate the possible combinations and arrangements

• What we are doing: Review the cases calculated above and find any error.
• Why: To ensure the answer is correct.

Let us consider the sum of the digits modulo 3: $4n_4 + 5n_5 + 9n_9 + 4 \equiv n_4 + 2n_5 + 0 + 1 \equiv 0 \pmod 3$. Therefore $n_4 + 2n_5 \equiv 2 \pmod 3$, where $n_4 + n_5 \le 5$.

When $n_5 = 0$, $n_4 \equiv 2 \pmod 3$, so $n_4 = 2, 5$. $n_4=2, n_5=0, n_9=3$: $\frac{5!}{2!0!3!} = 10$ $n_4=5, n_5=0, n_9=0$: $\frac{5!}{5!0!0!} = 1$ When $n_5 = 1$, $n_4 \equiv 0 \pmod 3$, so $n_4 = 0, 3$. $n_4=0, n_5=1, n_9=4$: $\frac{5!}{0!1!4!} = 5$ $n_4=3, n_5=1, n_9=1$: $\frac{5!}{3!1!1!} = 20$ When $n_5 = 2$, $n_4 \equiv 2-4 \equiv 1 \pmod 3$, so $n_4 = 1, 4$. $n_4=1, n_5=2, n_9=2$: $\frac{5!}{1!2!2!} = 30$ When $n_4=4, n_5=2, n_9=-1$, impossible When $n_5 = 3$, $n_4 \equiv 2-6 \equiv 2 \pmod 3$, so $n_4 = 2$. $n_4=2, n_5=3, n_9=0$: $\frac{5!}{2!3!0!} = 10$ When $n_5 = 4$, $n_4 \equiv 2-8 \equiv 0 \pmod 3$, so $n_4 = 0$. $n_4=0, n_5=4, n_9=1$: $\frac{5!}{0!4!1!} = 5$ When $n_5 = 5$, $n_4 \equiv 2-10 \equiv 1 \pmod 3$, so $n_4=-2$, impossible

The total number of combinations is $10 + 1 + 5 + 20 + 30 + 10 + 5 = 81$.

The correct answer is 81, which doesn't match the "Correct Answer" provided. There might be an error in the stated correct answer. After carefully checking the calculations, the answer should be 81.

Step 5: Double-check the problem statement and constraints

The problem states that the six-digit numbers are formed using the digits 4, 5, and 9 only. The number must be divisible by 6. The number of such numbers is 81.

Common Mistakes & Tips

• Always remember to consider all possible cases when dealing with divisibility rules.
• Be careful when calculating permutations with repeated elements.
• Double-check your calculations, especially when dealing with modular arithmetic.

Summary

To find the number of six-digit numbers formed using the digits 4, 5, and 9 that are divisible by 6, we first used the divisibility rule for 2 to determine that the last digit must be 4. Then, we used the divisibility rule for 3 to find the possible combinations of the first five digits such that the sum of all six digits is divisible by 3. Finally, we calculated the number of arrangements for each combination and summed them to get the total number of such six-digit numbers, which is 81.

Final Answer

The final answer is $\boxed{81}$.