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

m (→‎Solution: Author included 2013 in the first line, changed to explicity less than (as the problem implies).)
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==Solution==
 
==Solution==
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First, note that the only integer <math>2000\le x < 2013</math> is <math>2002</math>.
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We take cases on the thousands digit, which must be either <math>1</math> or <math>2</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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If the number is of the form <math>\overline{1bcd},</math> where <math>b, c, d</math> are digits, then we must have <math>d = 1 + b + c.</math> Since <math>d \le 9,</math> we must have <math>b + c \le 9 - 1 = 8.</math> By casework on the value of <math>b</math>, we find that there are <math>1 + 2 + \dots + 9 = 45</math> possible pairs <math>(b, c)</math>, and each pair uniquely determines the value of <math>d</math>, so we get <math>45</math> numbers with the given property.
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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>
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If the number is of the form <math>\overline{2bcd},</math> then it must be one of the numbers <math>2000, 2001, \dots, 2012.</math> Checking all these numbers, we find that only <math>2002</math> has the given property.
 +
Therefore, the number of numbers with the property is <math>45 + 1 = \boxed{46}.</math>
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2013|ab=B|num-b=17|num-a=19}}
 
{{AMC10 box|year=2013|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:23, 20 August 2020

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 $45 + 1 = \boxed{46}.$

See also

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
All AMC 10 Problems and Solutions

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