2021 AMC Fall 10A Problem 8

1. Write the number with its digits. Call the tens digit $a$ (from $1$ to $9$) and the units digit $b$ (from $0$ to $9$). For example, $47$ is $10 \times 4 + 7$. So the number is $10a + b$.
2. Write the rule as an equation. Cuddly means the number equals $a + b^2$:
$$10a + b = a + b^2.$$
3. Simplify. Subtract $a$ and $b$ from both sides:
$$9a = b^2 - b = b(b - 1).$$
So $b(b-1)$ must equal $9$ times the tens digit: one of $9, 18, 27, \ldots, 81$.
4. Check every units digit. There are only ten choices for $b$:

1. Only $72$ is on the list $9, 18, \ldots, 81$. It comes from $b = 9$, and $9a = 72$ gives $a = 8$. (The $0$s would need $a = 0$, which is not a tens digit.)
2. Check. $8 + 9^2 = 8 + 81 = 89$. It works.

There is exactly one cuddly number, $89$.

The answer is $\boxed{\textbf{(B)}\ 1}$.

Writing a two-digit number as $10a + b$ turns a word rule about digits into an equation. Here the equation leaves only ten cases, so checking them all is fastest and leaves no doubt. You could also notice that $b$ and $b - 1$ are next to each other, so they share no factor of $3$, which means one of them must be a multiple of $9$ by itself. That again points straight to $b = 9$.

Let the number have tens digit $a$ ($1 \le a \le 9$) and units digit $b$ ($0 \le b \le 9$), so its value is $10a + b$. Cuddly means
$$10a + b = a + b^2 \quad\Longrightarrow\quad 9a = b^2 - b = b(b-1).$$

So $b(b-1)$ must be a positive multiple of $9$ with $b \le 9$. The values of $b(b-1)$ for $b = 0, 1, \dots, 9$ are $0, 0, 2, 6, 12, 20, 30, 42, 56, 72$. The only multiples of $9$ are $0$ (from $b = 0$ or $1$) and $72$ (from $b = 9$). But $9a = 0$ forces $a = 0$, which is not allowed for a tens digit. The remaining case $b = 9$ gives $9a = 72$, so $a = 8$.

The unique cuddly number is $89$; indeed $8 + 9^2 = 89$.

The answer is $\boxed{\textbf{(B)}\ 1}$.