1. Form of the number

Let the 3-digit odd number be $\overline{abc}$, where:

• $a \in \{1,2,\dots,9\}$ (hundreds digit, nonzero),
• $b \in \{0,1,2,\dots,9\}$,
• $c \in \{1,3,5,7,9\}$ (units digit, since the number is odd).

We need: $a+b+c \equiv 0 \pmod{7}.$

So we count all triples $(a,b,c)$ satisfying this.

2. Possible sums of digits

Since

• minimum sum $=1+0+1=2$,
• maximum sum $=9+9+9=27$,

the multiples of $7$ possible are: $7,14,21.$

So we count 3-digit odd numbers with digit sum equal to $7$, $14$, or $21$.

3. Count solutions for each odd unit digit

Fix the unit digit $c \in \{1,3,5,7,9\}$. Then we need: $a+b = 7-c,\; 14-c,\; 21-c.$

with constraints: $1 \le a \le 9, \qquad 0 \le b \le 9.$

For a fixed value $a+b=n$, the number of solutions is the number of integers $a$ such that $1 \le a \le 9, \qquad 0 \le n-a \le 9.$

Equivalently, $\max(1,n-9) \le a \le \min(9,n).$

Now compute casewise.

Case 1: $c=1$

Then $a+b=6,13,20.$

• For $a+b=6$: $a=1$ to $6$ gives $6$ solutions.
• For $a+b=13$: $a=4$ to $9$ gives $6$ solutions.
• For $a+b=20$: impossible since max $a+b=18$.

Total for $c=1$: $6+6=12.$

Case 2: $c=3$

Then $a+b=4,11,18.$

• For $a+b=4$: $a=1$ to $4$ gives $4$ solutions.
• For $a+b=11$: $a=2$ to $9$ gives $8$ solutions.
• For $a+b=18$: only $(a,b)=(9,9)$, so $1$ solution.

Total for $c=3$: $4+8+1=13.$

Case 3: $c=5$

Then $a+b=2,9,16.$

• For $a+b=2$: $(a,b)=(1,1),(2,0)$, so $2$ solutions.
• For $a+b=9$: $a=1$ to $9$, so $9$ solutions.
• For $a+b=16$: $a=7,8,9$ gives $3$ solutions.

Total for $c=5$: $2+9+3=14.$

Case 4: $c=7$

Then $a+b=0,7,14.$

• For $a+b=0$: impossible since $a\ge1$.
• For $a+b=7$: $a=1$ to $7$ gives $7$ solutions.
• For $a+b=14$: $a=5$ to $9$ gives $5$ solutions.

Total for $c=7$: $0+7+5=12.$

Case 5: $c=9$

Then $a+b=-2,5,12.$

• For $a+b=-2$: impossible.
• For $a+b=5$: $a=1$ to $5$ gives $5$ solutions.
• For $a+b=12$: $a=3$ to $9$ gives $7$ solutions.

Total for $c=9$: $0+5+7=12.$

4. Add all cases

Total number of required 3-digit odd numbers: $12+13+14+12+12=63.$

5. Comparison with stored answer

Our derived answer is: $\boxed{63}$

This matches the stored correct answer.