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

First, note that the only integer $2000\le x \le 2013$ is $2002$ . Now let's look at all numbers $x$ where $1000<x<2000$ Let the hundreds digit be $0$ . Then, the tens and units digit can be $01, 12, 23, \hdots, 89$ , which is $9$ possibilities. We notice as the hundreds digit goes up by one, the number of possibilities goes down by one. Thus, the number of integers is $1+2+3+4+5+6+7+8+9+1=\boxed{\textbf{(D)} 46}$

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2013 AMC 10B Problems/Problem 18

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Solution

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