2021 AMC Fall 10A Problem 8

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 $\boxed{\textbf{(B)}\ 1}$.

Digit problems become algebra once you write the number as $10a + b$. Here the equation forces a divisibility condition, $9 \mid b(b-1)$, and with only ten possible units digits a direct check is fastest. Note that $b$ and $b - 1$ are consecutive, so $9$ has to divide one of them entirely, which is why only $b = 9$ (and the degenerate $b = 0, 1$) appear.

• Counting b = 0 or b = 1 (which give a = 0) as solutions, even though the tens digit must be nonzero.
• Writing the number as $a + b$ or squaring the tens digit instead of the units digit.