2001 AMC 10 Problem 6

P1-10~2 minPrint

Bases & Digits Set up the equation/formula and compute; no special trick needed

Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits of the integer $n$ . For example, $P(23) = 6$ and $S(23) = 5$ . Suppose $N$ is a two-digit number such that $N = P(N)+S(N)$ . What is the units digit of $N$ ?

Show solution without answering

Key insight

Write the number as 10 times the tens digit plus the units digit. The units digit cancels, and what is left forces it to be 9.

Solution

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

Why this works

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.

Alternative approach

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 $9$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Number Theory, Chapter 9: Base Number Arithmetic
Introduction to Number Theory, Chapter 6: Special Numbers (§6.5)