Difference between revisions of "2013 AMC 10B Problems/Problem 18"
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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. | 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 == | ||
| + | {{AMC10 box|year=2013|ab=B|num-b=17|num-a=19}} | ||
Revision as of 17:03, 27 March 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 
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 | ||
