Difference between revisions of "2013 AMC 10B Problems/Problem 18"
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==Solution== | ==Solution== | ||
| + | First, note that the only integer <math>2000\le x \le 2013</math> is <math>2002</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. | ||
| + | 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 21:49, 21 February 2013
Problem
The number 
has the property that its units digit is the sum of its other digits, that is 
. How many integers less than but greater than 
share this property?
Solution
First, note that the only integer 
is 
.
Now let's look at all numbers 
where 
Let the hundreds digit be 
. Then, the tens and units digit can be 
, which is 
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 
