2013 AMC 10B Problem 18

P16-20~4 minPrint

Bases & Digits Basic Counting Split into exhaustive cases and handle each Set up the equation/formula and compute; no special trick needed

The number $2013$ has the property that its units digit is the sum of its other digits, that is $2+0+1=3$ . How many integers less than $2013$ but greater than $1000$ share this property?

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Key insight

For numbers 1bcd, the last digit is 1 + b + c, so b + c is at most 8: that gives 45 numbers. Add 2002.

Solution

The numbers run from $1001$ to $2012$ . Split them into the ones starting with $1$ and the ones starting with $2$ .

Numbers from $1001$ to $1999$ .

Write such a number with digits $1, b, c, d$ . We need $d = 1 + b + c$ .
Once $b$ and $c$ are chosen, $d$ is fixed. The only catch is that $d$ must be a single digit, at most $9$ , so $b + c$ can be at most $8$ .
Count the pairs $(b, c)$ for each possible sum:

$b + c$	0	1	2	3	4	5	6	7	8
number of pairs	1	2	3	4	5	6	7	8	9

(For example, sum $2$ has the pairs $(0,2), (1,1), (2,0)$ .) The total is $1 + 2 + \cdots + 9 = 45$ .
4. The pair $b = c = 0$ gives $1001$ , which is more than $1000$ , so it counts.

Numbers from $2000$ to $2012$ .

These have digits $2, 0, c, d$ with $c = 0$ or $1$ . We need $d = 2 + c$ . With $c = 0$ we get $2002$ , which works. With $c = 1$ we get $2013$ , which is not less than $2013$ , so it is out. That is $1$ more number.

Total: $45 + 1 = 46$ .

The answer is $\boxed{\textbf{(D)}\ 46}$ .

Why this works

Once the first digit is fixed, the last digit is decided by the middle two, so the problem becomes counting pairs of digits with a limited sum, which gives $1 + 2 + \cdots + 9$ . The upper limit $2013$ is a deliberate edge: always check the short, awkward range at the end separately.

Concise solution

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

Write the number as $\overline{abcd}$ with $a \in \{1, 2\}$ and the condition $d = a + b + c$ , which requires $a + b + c \le 9$ .

Thousands digit $1$ . Numbers $1001$ to $1999$ : we need $d = 1 + b + c$ , so $b + c \le 8$ , and each such pair $(b, c)$ determines $d$ uniquely. For each sum $s = b + c$ from $0$ to $8$ there are $s + 1$ pairs, so the count is
$1 + 2 + \cdots + 9 = 45.$
(The pair $b = c = 0$ gives $1001$ , which is greater than $1000$ and allowed.)

Thousands digit $2$ . Only $2000$ to $2012$ are in range, so $b = 0$ and $c \in \{0, 1\}$ . With $c = 0$ , $d = 2$ : $2002$ works. With $c = 1$ , $d = 3$ gives $2013$ , which is excluded. That is $1$ more.

Total: $45 + 1 = 46$ .

The answer is $\boxed{\textbf{(D)}\ 46}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Number Theory, Chapter 9: Base Number Arithmetic