Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

(→‎See also: Removed spam.)
m (→‎Solution: Author included 2013 in the first line, changed to explicity less than (as the problem implies).)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
First, note that the only integer <math>2000\le x \le 2013</math> is <math>2002</math>.
+
First, note that the only integer <math>2000\le x < 2013</math> is <math>2002</math>.
 
Now let's look at all numbers <math>x</math> where <math>1000<x<2000</math>
 
Now let's look at all numbers <math>x</math> where <math>1000<x<2000</math>
 
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.
 
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.
 
Thus, the number of integers is <math>1+2+3+4+5+6+7+8+9+1=\boxed{\textbf{(D)} 46}</math>
 
Thus, the number of integers is <math>1+2+3+4+5+6+7+8+9+1=\boxed{\textbf{(D)} 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 13:04, 3 February 2015

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 $2000\le x < 2013$ 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 $1+2+3+4+5+6+7+8+9+1=\boxed{\textbf{(D)} 46}$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_18&oldid=67243"