Difference between revisions of "1983 AIME Problems/Problem 10"
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Suppose the two identical [[digit]]s are both one. Since the thousands digits must be one, the other one can be in only one of three digits, | Suppose the two identical [[digit]]s are both one. Since the thousands digits must be one, the other one can be in only one of three digits, | ||
| − | <center><math>11xy,\qquad 1x1y,\qquad1xy1.</math></ | + | <div style="text-align:center;"><math>11xy,\qquad 1x1y,\qquad1xy1.</math></div> |
| − | Because the number must have exactly two identical digits, <math>x\neq y</math>, <math>x\neq1</math>, and <math>y\neq1</math>. Hence, there are <math>3\cdot9\cdot8=216</math> numbers of this form. | + | Because the number must have exactly two identical digits, <math>\displaystyle x\neq y</math>, <math>\displaystyle x\neq1</math>, and <math>\displaystyle y\neq1</math>. Hence, there are <math>3\cdot9\cdot8=216</math> numbers of this form. |
Suppose the two identical digits are not one. Therefore, consider the following possibilities, | Suppose the two identical digits are not one. Therefore, consider the following possibilities, | ||
| − | <center><math>1xxy,\qquad1xyx,\qquad1yxx.</math></ | + | <div style="text-align:center;"><math>1xxy,\qquad1xyx,\qquad1yxx.</math></div> |
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 too. | 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 too. | ||
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Thus, the desired answer is <math>216+216=432</math>. | Thus, the desired answer is <math>216+216=432</math>. | ||
| − | + | == See also == | |
| − | |||
{{AIME box|year=1983|num-b=9|num-a=11}} | {{AIME box|year=1983|num-b=9|num-a=11}} | ||
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* [[AIME Problems and Solutions]] | * [[AIME Problems and Solutions]] | ||
* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] | ||
Revision as of 17:40, 21 March 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 
.
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 | ||
