This question asks us to find the total number of three-digit PINs where the third digit is the sum of the first two digits. A three-digit PIN can be represented as $d_1 d_2 d_3$, where each $d_i$ is a digit from 0 to 9.

Understanding the PIN Condition

Let the three digits of the PIN be $d_1$, $d_2$, and $d_3$. The problem states the condition that the third digit must be the sum of the first two digits. Mathematically, this is represented as:

$ d_3 = d_1 + d_2 $

Since $d_3$ must be a single digit, it must be less than or equal to 9. Therefore, the condition becomes:

$ d_1 + d_2 \le 9 $

We need to consider all possible combinations of $d_1$ and $d_2$ (where $d_1, d_2 \in \{0, 1, ..., 9\}$) that satisfy this inequality.

Counting Valid PIN Combinations

We can count the valid PINs by iterating through the possible values for the first digit, $d_1$, and determining the number of valid choices for the second digit, $d_2$, for each $d_1$. The third digit, $d_3$, is then fixed by the sum.

• If $d_1 = 0$, then $d_2$ can be any digit from 0 to 9 ($0+d_2 \le 9$). This gives 10 possibilities (000, 011, ..., 099).
• If $d_1 = 1$, then $d_2$ can be any digit from 0 to 8 ($1+d_2 \le 9$). This gives 9 possibilities (101, 112, ..., 189).
• If $d_1 = 2$, then $d_2$ can be any digit from 0 to 7 ($2+d_2 \le 9$). This gives 8 possibilities (202, 213, ..., 279).
• If $d_1 = 3$, then $d_2$ can be any digit from 0 to 6 ($3+d_2 \le 9$). This gives 7 possibilities.
• If $d_1 = 4$, then $d_2$ can be any digit from 0 to 5 ($4+d_2 \le 9$). This gives 6 possibilities.
• If $d_1 = 5$, then $d_2$ can be any digit from 0 to 4 ($5+d_2 \le 9$). This gives 5 possibilities.
• If $d_1 = 6$, then $d_2$ can be any digit from 0 to 3 ($6+d_2 \le 9$). This gives 4 possibilities.
• If $d_1 = 7$, then $d_2$ can be any digit from 0 to 2 ($7+d_2 \le 9$). This gives 3 possibilities.
• If $d_1 = 8$, then $d_2$ can be 0 or 1 ($8+d_2 \le 9$). This gives 2 possibilities.
• If $d_1 = 9$, then $d_2$ must be 0 ($9+d_2 \le 9$). This gives 1 possibility (909).

Calculating the Total Number of PINs

To find the total number of valid three-digit PINs, we sum the possibilities for each value of $d_1$:

Total PINs = $10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1$

This is the sum of the first 10 positive integers. Using the formula for the sum of an arithmetic series, $S_n = \frac{n(n+1)}{2}$, where $n=10$:

Total PINs = $\sum_{i=0}^{9} (10-i) = \frac{10 \times (10+1)}{2} = \frac{10 \times 11}{2} = 55$

Therefore, there are 55 possible three-digit PINs where the third digit is the sum of the first two digits.