Difference between revisions of "1983 AIME Problems/Problem 10"
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== See also == | == See also == | ||
Revision as of 20:53, 1 February 2007
Problem
The numbers 
, 
, and 
have something in common. Each is a four-digit number beginning with 
that has exactly two identical digits. How many such numbers are there?
Solution
Suppose the two identical digits are both one. Since the thousands digits must be one, the other one can be in only one of three digits,
Because the number must have exactly two identical digits, 
, 
, and 
. Hence, there are 
numbers of this form.
Suppose the two identical digits are not one. Therefore, consider the following possibilities,
Again, , 
, and 
. There are numbers of this form too.
Thus, the desired answer is 
.
| 1983 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
