Key Concepts and Formulas

• A 3-digit number can be represented as $100x + 10y + z$, where $x$ is a digit from 1 to 9, and $y$ and $z$ are digits from 0 to 9.
• For a number to be odd, its units digit ($z$) must be one of $\{1, 3, 5, 7, 9\}$.
• The sum of the digits is $x+y+z$, which must be a multiple of 7, i.e., $x+y+z = 7k$ for some integer $k$.

Step-by-Step Solution

Step 1: Analyze the possible values for the units digit $z$. Since the number must be odd, $z$ can be 1, 3, 5, 7, or 9. We will consider each case separately.

Step 2: Case 1: $z = 1$. We need $x + y + 1 = 7k$, which implies $x + y = 7k - 1$. Since $x$ is between 1 and 9 inclusive, and $y$ is between 0 and 9 inclusive, the minimum value of $x+y$ is $1+0=1$ and the maximum value is $9+9=18$. Therefore, $7k-1$ can take values 7 and 14.

• If $x+y = 6$, the possible pairs of $(x, y)$ are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 0)$. There are 6 solutions.
• If $x+y = 13$, the possible pairs of $(x, y)$ are $(4, 9), (5, 8), (6, 7), (7, 6), (8, 5), (9, 4)$. There are 6 solutions. Total solutions for this case: $6+6 = 12$.

Step 3: Case 2: $z = 3$. We need $x + y + 3 = 7k$, which implies $x + y = 7k - 3$. Since $x+y$ is between 1 and 18, $7k-3$ can take values 4, 11, and 18.

• If $x+y = 4$, the possible pairs of $(x, y)$ are $(1, 3), (2, 2), (3, 1), (4, 0)$. There are 4 solutions.
• If $x+y = 11$, the possible pairs of $(x, y)$ are $(2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)$. There are 8 solutions.
• If $x+y = 18$, the possible pairs of $(x, y)$ are $(9, 9)$. There is 1 solution. Total solutions for this case: $4+8+1 = 13$.

Step 4: Case 3: $z = 5$. We need $x + y + 5 = 7k$, which implies $x + y = 7k - 5$. Since $x+y$ is between 1 and 18, $7k-5$ can take values 2, 9, and 16.

• If $x+y = 2$, the possible pairs of $(x, y)$ are $(1, 1), (2, 0)$. There are 2 solutions.
• If $x+y = 9$, the possible pairs of $(x, y)$ are $(1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0)$. There are 9 solutions.
• If $x+y = 16$, the possible pairs of $(x, y)$ are $(7, 9), (8, 8), (9, 7)$. There are 3 solutions. Total solutions for this case: $2+9+3 = 14$.

Step 5: Case 4: $z = 7$. We need $x + y + 7 = 7k$, which implies $x + y = 7k - 7$. Since $x+y$ is between 1 and 18, $7k-7$ can take values 0, 7, and 14.

• If $x+y = 0$, there are no solutions since $x$ must be at least 1. So there are 0 solutions.
• If $x+y = 7$, the possible pairs of $(x, y)$ are $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (7, 0)$. There are 7 solutions.
• If $x+y = 14$, the possible pairs of $(x, y)$ are $(5, 9), (6, 8), (7, 7), (8, 6), (9, 5)$. There are 5 solutions. Total solutions for this case: $0+7+5 = 12$.

Step 6: Case 5: $z = 9$. We need $x + y + 9 = 7k$, which implies $x + y = 7k - 9$. Since $x+y$ is between 1 and 18, $7k-9$ can take values 5, 12, and 19.

• If $x+y = 5$, the possible pairs of $(x, y)$ are $(1, 4), (2, 3), (3, 2), (4, 1), (5, 0)$. There are 5 solutions.
• If $x+y = 12$, the possible pairs of $(x, y)$ are $(3, 9), (4, 8), (5, 7), (6, 6), (7, 5), (8, 4), (9, 3)$. There are 7 solutions.
• If $x+y = 19$, there are no solutions since the maximum value of $x+y$ is $9+9 = 18$. So there are 0 solutions. Total solutions for this case: $5+7+0 = 12$.

Step 7: Calculate the total number of such 3-digit odd numbers. Total number of solutions = $12 + 13 + 14 + 12 + 12 = 63$.

Common Mistakes & Tips

• Remember that the first digit of a 3-digit number cannot be 0.
• Be careful when listing pairs of digits whose sum is a certain value. Make sure you don't miss any pairs or include any invalid pairs.
• Double-check your arithmetic when summing the number of solutions for each case.

Summary

We analyzed the possible values for the units digit of the 3-digit odd number and considered each case separately. For each case, we found the possible pairs of digits for the hundreds and tens places such that the sum of all three digits is a multiple of 7. Then we summed the number of solutions from each case to find the total number of such 3-digit odd numbers.

The final answer is \boxed{63}.