To find the number of 3-digit odd numbers whose sum of digits is a multiple of 7, we proceed as follows:
Let the 3-digit number be represented as \( \overline{abc} \), where \( a, b, c \) are digits. Since it's an odd number, \( c \) must be odd: \( c \in \{1, 3, 5, 7, 9\} \).
The number must also satisfy \( a + b + c \) being a multiple of 7. We have the following constraints: \( a \geq 1 \) (since it's a 3-digit number) and \( 0 \leq b \leq 9 \).
We want to find all combinations where \( a + b + c \equiv 0 \mod 7 \). Iterate over possible values of \( c \) and solve for \( a + b \):
For \( c = 1, a + b \equiv 6 \mod 7 \). This implies possibilities such as \( 1+5, 2+4, 3+3, 4+2, 5+1, 6+0 \).
For \( c = 3, a + b \equiv 4 \mod 7 \). This implies possibilities such as \( 1+3, 2+2, 3+1, 4+0 \).
For \( c = 5, a + b \equiv 2 \mod 7 \). This implies possibilities such as \( 1+1, 2+0 \).
For \( c = 7, a + b \equiv 0 \mod 7 \). This implies possibility: \( 1+0 \).
For \( c = 9, a + b \equiv 5 \mod 7 \). This implies possibilities such as \( 2+3, 3+2, 4+1, 5+0 \).
Count the valid combinations for each case:
For \( c = 1 \), valid \( (a, b) \) are \( 15, 24, 33, 42, 51, 60 \) (6 combinations).
For \( c = 3 \), valid \( (a, b) \) are \( 13, 22, 31, 40 \) (4 combinations).
For \( c = 5 \), valid \( (a, b) \) are \( 11, 20 \) (2 combinations).
For \( c = 7 \), valid \( (a, b) \) is \( 10 \) (1 combination).
For \( c = 9 \), valid \( (a, b) \) are \( 23, 32, 41, 50 \) (4 combinations).
