2013 AMC 10B Problems/Problem 18

Problem

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?

$\textbf{(A)}\ 33\qquad\textbf{(B)}\ 34\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ 58$

Solution

We take cases on the thousands digit, which must be either $1$ or $2$ : If the number is of the form $\overline{1bcd},$ where $b, c, d$ are digits, then we must have $d = 1 + b + c.$ Since $d \le 9,$ we must have $b + c \le 9 - 1 = 8.$ By casework on the value of $b$ , we find that there are $1 + 2 + \dots + 9 = 45$ possible pairs $(b, c)$ , and each pair uniquely determines the value of $d$ , so we get $45$ numbers with the given property.

If the number is of the form $\overline{2bcd},$ then it must be one of the numbers $2000, 2001, \dots, 2012.$ Checking all these numbers, we find that only $2002$ has the given property. Therefore, the number of numbers with the property is $45 + 1 = \boxed{46}.$

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2013 AMC 10B Problems/Problem 18

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