Key Concepts and Formulas

• Divisibility by 15: A number is divisible by 15 if and only if it is divisible by both 3 and 5, since 15 = 3 x 5 and 3 and 5 are coprime.
• Divisibility Rule for 5: A number is divisible by 5 if and only if its last digit is 0 or 5.
• Divisibility Rule for 3: A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
• Permutations: The number of ways to arrange $n$ distinct objects in $r$ places is given by $P(n, r) = \frac{n!}{(n-r)!}$.

Step-by-Step Solution

Step 1: Apply the Divisibility Rule for 5 Since the 6-digit number must be divisible by 5, its last digit must be either 0 or 5. However, we can only use the digits 1, 2, 3, 4, 5, 6, and 7. Therefore, the last digit must be 5.

Step 2: Apply the Divisibility Rule for 3 Since the 6-digit number must be divisible by 3, the sum of its digits must be divisible by 3. The digits we are using are from the set $\{1, 2, 3, 4, 5, 6, 7\}$. Since the last digit is 5, we need to choose 5 more digits such that their sum, plus 5, is divisible by 3. The sum of all the digits is $1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$. Let $S$ be the sum of the 5 digits we choose (excluding 5). We need $S + 5$ to be divisible by 3, which means $S + 5 \equiv 0 \pmod{3}$, or $S \equiv 1 \pmod{3}$.

Step 3: Determine which digit to exclude We are choosing 5 digits from the set $\{1, 2, 3, 4, 6, 7\}$. The sum of these 6 digits is $1+2+3+4+6+7 = 23$. Let $x$ be the digit that we exclude. Then the sum of the 5 digits we use is $S = 23 - x$. We need $S \equiv 1 \pmod{3}$, so $23 - x \equiv 1 \pmod{3}$. This simplifies to $2 - x \equiv 1 \pmod{3}$, or $x \equiv 1 \pmod{3}$. The digits in the set $\{1, 2, 3, 4, 6, 7\}$ that satisfy $x \equiv 1 \pmod{3}$ are 1, 4, and 7.

Step 4: Analyze the possible sets of digits We have three cases:

• Case 1: Exclude 1. The digits are $\{2, 3, 4, 6, 7, 5\}$.
• Case 2: Exclude 4. The digits are $\{1, 2, 3, 6, 7, 5\}$.
• Case 3: Exclude 7. The digits are $\{1, 2, 3, 4, 6, 5\}$.

In each case, we have a set of 6 distinct digits. The last digit must be 5, so we have 5 remaining digits to arrange in the first 5 places. The number of ways to arrange 5 distinct digits in 5 places is $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

Step 5: Calculate the total number of 6-digit numbers Since there are 3 possible sets of digits, and each set can form $5! = 120$ numbers, the total number of such 6-digit numbers is $3 \times 120 = 360$.

Common Mistakes & Tips

• Remember to check both divisibility rules (3 and 5).
• Carefully consider which digit is fixed (in this case, the last digit).
• Don't forget to calculate the number of permutations correctly.

Summary

To form a 6-digit number using the digits 1, 2, 3, 4, 5, 6, 7 without repetition that is divisible by 15, we must ensure it is divisible by both 3 and 5. This means the last digit must be 5, and the sum of the digits must be divisible by 3. We identified the three sets of digits that satisfy these conditions, and for each set, we calculated the number of permutations with 5 as the last digit. The total number of such numbers is 360.

Final Answer

The final answer is \boxed{360}.