Difference between revisions of "1983 AIME Problems/Problem 10"
m |
m |
||
| Line 15: | Line 15: | ||
Again, <math>x\neq y</math>, <math>x\neq 1</math>, and <math>y\neq 1</math>. There are <math>3\cdot9\cdot8=216</math> numbers of this form as well. | Again, <math>x\neq y</math>, <math>x\neq 1</math>, and <math>y\neq 1</math>. There are <math>3\cdot9\cdot8=216</math> numbers of this form as well. | ||
| − | Thus, the desired answer is <math>216+216=432</math>. | + | Thus, the desired answer is <math>216+216=\boxed{432}</math>. |
== See Also == | == See Also == | ||
Revision as of 06:00, 5 December 2013
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 as well.
Thus, the desired answer is 
.
See Also
| 1983 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
