2001 AMC 10 Problem 6

1. Try one number to see what the condition means. For $N = 23$: $P + S = 6 + 5 = 11$, which is not $23$. So most numbers fail.
2. To see which ones work, call the tens digit $a$ and the units digit $b$. Then $N = 10a + b$ (for example $23 = 10 \cdot 2 + 3$).
3. The product of the digits is $ab$ and the sum is $a + b$. The condition $N = P(N) + S(N)$ becomes
$$10a + b = ab + a + b.$$
4. Both sides have a $+\,b$, so take it away from both sides:
$$10a = ab + a.$$
5. Take $a$ away from both sides: $9a = ab$.
6. The tens digit $a$ is not $0$ (otherwise $N$ would not be a two-digit number), so we can divide both sides by $a$. That gives $b = 9$.
7. Check with $N = 19$: $P = 9$, $S = 10$, and $9 + 10 = 19$. It works. So do $29, 39, \ldots, 99$, and every one ends in $9$.

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

Writing a two-digit number as $10a + b$ turns a digit puzzle into ordinary algebra. Here the units digit shows up on both sides and cancels, which is why the question asks only about the units digit: the tens digit can be anything. Before dividing by a letter, always make sure it cannot be $0$; a leading digit never is.

Test small numbers, starting with those ending in the answer choices. $12$: $2 + 3 = 5$, no. $13$: $3 + 4 = 7$, no. $16$: $6 + 7 = 13$, no. $18$: $8 + 9 = 17$, no. $19$: $9 + 10 = 19$, yes. The units digit is $9$.

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 $\boxed{\textbf{(E)}\ 9}$.