2001 AMC 10 Problem 6

Let the tens digit be $a$ and the units digit be $b$, so $N = 10a + b$ with $a \ge 1$. Then $P(N) = ab$ and $S(N) = a + b$, and the condition reads
$$10a + b = ab + a + b.$$
The $b$ on each side cancels, leaving $10a = ab + a$, i.e. $9a = ab$. Because $a \neq 0$ (it is the leading digit), divide by $a$ to get $b = 9$.

Every number $19, 29, \dots, 99$ works (for instance $19 = 9 + 10$), and in all of them the units digit is $9$.

The answer is $\boxed{\textbf{(E)}\ 9}$.

Digit conditions become algebra as soon as you write the number in place-value form. Here the units digit appears on both sides and cancels, so the tens digit is unconstrained; the problem asks for the units digit precisely because it is the only thing determined. Watch for a leading digit that is guaranteed nonzero, which licenses the division.

Try small numbers: $19$ gives $P + S = 9 + 10 = 19$, a hit. Its units digit is $9$; since the problem promises a single answer, choose (E).

• Dividing by a without noting a is nonzero, or searching for a specific N (like 19) and reporting its tens digit instead of the units digit.
• Mixing up $P$ and $S$, or setting $N = ab + a + b$ with $N = a + b$ instead of $10a + b$.