2016 AMC 10B Problem 24

Let the common difference be $D = 10p + q$ with $0 \le q \le 9$. Adding $D$ to $\overline{ab}$ must produce $\overline{bc}$, and adding it to $\overline{bc}$ must produce $\overline{cd}$.

Track the digits. In the first addition, the units digits satisfy $c \equiv b + q \pmod{10}$ with a carry $k_1 \in \{0, 1\}$, and the tens digits satisfy $b = a + p + k_1$. In the second addition, $d \equiv c + q \pmod{10}$ with carry $k_2$, and $c = b + p + k_2$.

The step from $b$ to $c$ appears twice: as units digits, $c = b + q - 10k_1$; as tens digits, $c = b + p + k_2$. Hence $q - 10k_1 = p + k_2 \ge 0$, which forces $k_1 = 0$ and $q = p + k_2$. So $D = 11p + k_2$.

Case $k_2 = 0$: $D = 11p$, no carries, and $a, b, c, d$ increase by $p \ge 1$ each step. Need $d = a + 3p \le 9$ with $a \ge 1$: $p = 1$ gives $a = 1, \ldots, 6$ ($6$ numbers such as $1234$), $p = 2$ gives $a = 1, 2, 3$ ($3$ numbers such as $2468$). That is $9$.

Case $k_2 = 1$: $D = 11p + 1$, so $b = a + p$, $c = b + p + 1$, and $d = c + p + 1 - 10$. Constraints: $c = a + 2p + 1 \le 9$ and $d \ge 0$, i.e. $a + 3p + 2 \ge 10$. So $8 - 3p \le a \le 8 - 2p$ with $a \ge 1$:
- $p = 0$: $a = 8$, giving $8890$ ($88, 89, 90$).
- $p = 1$: $a = 5, 6$, giving $5680, 6791$.
- $p = 2$: $a = 2, 3, 4$, giving $2470, 3581, 4692$.
- $p = 3$: $a = 1, 2$, giving $1482, 2593$.
That is $1 + 2 + 3 + 2 = 8$.

Total: $9 + 8 = 17$.

The answer is $\boxed{\textbf{(D)}\ 17}$.

The overlapping digits ($b$ is both a units digit and a tens digit) let you compare two descriptions of the same step, which pins down the common difference modulo $11$ and rules out a carry in the first addition. After that the problem is two small, fully determined cases. In digit problems, write each addition column by column with explicit carries rather than reasoning about the numbers as wholes.

Bound and search: $D = \overline{bc} - \overline{ab}$ and $2D = \overline{cd} - \overline{ab} \le 98 - 10$, so $D \le 44$. Then $D \equiv 0$ or $1 \pmod{11}$ leaves $D \in \{1, 11, 12, 22, 23, 33, 34, 44\}$, and a quick check of each (for instance $D = 33$ and $D = 44$ give nothing) reproduces the $17$ numbers.

• Counting only the carry-free numbers with digits in arithmetic progression ($1234$, $2468$, ...), which gives $9$ (choice A) and misses cases like $4692$.
• Rejecting $d = 0$ (as in $8890$ or $5680$); only $a$ is required to be nonzero, and $\overline{cd}$ is still a two-digit number.