Difference between revisions of "2013 AMC 10B Problems/Problem 18"

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

Revision as of 22:49, 21 February 2013

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 $1+2+3+4+5+6+7+8+9+1=\boxed{\textbf{(D)} 46}$