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

m (See also)
(See also)
Line 12: Line 12:
 
{{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}}
This is right.
+
This is right. Difficulty Level is Okay....
 +
/\/\/\
 +
|>.<|
 +
| --- |
 +
\__/

Revision as of 00:36, 26 January 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 $2000\le x \le 2013$ 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 $1+2+3+4+5+6+7+8+9+1=\boxed{\textbf{(D)} 46}$

See also

2013 AMC 10B (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

This is right. Difficulty Level is Okay.... /\/\/\ |>.<| | --- |

\__/