Key Concepts and Formulas

• Divisibility Rule for 6: A number is divisible by 6 if and only if it is divisible by both 2 and 3.
• 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.

Step-by-Step Solution

Step 1: Applying the Divisibility Rule for 2

Since the 3-digit number must be divisible by 6, it must also be divisible by 2. Therefore, the last digit must be even. The possible digits are {1, 2, 3, 4, 5}, so the last digit can only be 2 or 4. This gives us two cases:

• Case 1: The number is of the form $ab2$.
• Case 2: The number is of the form $ab4$.

Step 2: Analyzing Case 1: Units Digit $c = 2$

We have a number of the form $ab2$. For this number to be divisible by 3, the sum of its digits, $a + b + 2$, must be divisible by 3. This means $a + b + 2 \equiv 0 \pmod{3}$, or $a + b \equiv -2 \pmod{3}$. Since $-2 \equiv 1 \pmod{3}$, we have $a + b \equiv 1 \pmod{3}$. The digits $a$ and $b$ can be any of {1, 2, 3, 4, 5}.

The possible values for $a+b$ are 4, 7, and 10, because these are the only numbers in the range of possible sums ($1+1=2$ to $5+5=10$) that have a remainder of 1 when divided by 3.

• If $a + b = 4$, the possible pairs $(a, b)$ are (1, 3), (2, 2), (3, 1).
• If $a + b = 7$, the possible pairs $(a, b)$ are (2, 5), (3, 4), (4, 3), (5, 2).
• If $a + b = 10$, the possible pair $(a, b)$ is (5, 5).

Therefore, there are $3 + 4 + 1 = 8$ such numbers in Case 1.

Step 3: Analyzing Case 2: Units Digit $c = 4$

We have a number of the form $ab4$. For this number to be divisible by 3, the sum of its digits, $a + b + 4$, must be divisible by 3. This means $a + b + 4 \equiv 0 \pmod{3}$, or $a + b \equiv -4 \pmod{3}$. Since $-4 \equiv 2 \pmod{3}$, we have $a + b \equiv 2 \pmod{3}$. The digits $a$ and $b$ can be any of {1, 2, 3, 4, 5}.

The possible values for $a+b$ are 2, 5, and 8, because these are the only numbers in the range of possible sums ($1+1=2$ to $5+5=10$) that have a remainder of 2 when divided by 3.

• If $a + b = 2$, the possible pair $(a, b)$ is (1, 1).
• If $a + b = 5$, the possible pairs $(a, b)$ are (1, 4), (2, 3), (3, 2), (4, 1).
• If $a + b = 8$, the possible pairs $(a, b)$ are (3, 5), (4, 4), (5, 3).

Therefore, there are $1 + 4 + 3 = 8$ such numbers in Case 2.

Step 4: Calculating the Total Number of 3-Digit Numbers

The total number of 3-digit numbers that are divisible by 6 and meet all the given criteria is the sum of the counts from Case 1 and Case 2.

Total numbers = (Numbers ending in 2) + (Numbers ending in 4) Total numbers = $8 + 8 = 16$.

Common Mistakes & Tips

• Forgetting Repetition: Remember that repetition of digits is allowed. This means digits like $a=b$ (e.g., 222, 114, 552) are valid.
• Systematic Listing: When listing pairs $(a,b)$, it's helpful to fix one digit (e.g., $a$) and then find corresponding $b$ values, or simply list them in ascending order of $a$ to ensure no combinations are missed and no duplicates are counted.
• Using Modular Arithmetic: Simplify divisibility conditions using congruences.

Summary

To find the total number of 3-digit numbers divisible by 6 formed by the digits {1, 2, 3, 4, 5} with repetition, we first enforced divisibility by 2, restricting the last digit to 2 or 4. Then, for each case, we enforced divisibility by 3 by considering the possible sums of the first two digits modulo 3. Finally, summing the counts from both cases, we find the total number of such numbers.

The final answer is $\boxed{16}$.