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Difference between revisions of "1983 AIME Problems/Problem 10"

 
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== Problem ==
 
== Problem ==
 +
The numbers <math>1447</math>, <math>1005</math>, and <math>1231</math> have something in common. Each is a four-digit number beginning with <math>1</math> that has exactly two identical digits. How many such numbers are there?
  
 
== Solution ==
 
== Solution ==
 +
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,
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<center><math>11xy,\qquad 1x1y,\qquad1xy1.</math></center>
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 +
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.
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 +
Suppose the two identical digits are not one. Therefore, consider the following possibilities,
 +
 +
<center><math>1xxy,\qquad1xyx,\qquad1yxx.</math></center>
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 +
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.
 +
 +
Thus, the desired answer is <math>216+216=432</math>.
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----
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* [[1983 AIME Problems/Problem 9|Previous Problem]]
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* [[1983 AIME Problems/Problem 11|Next Problem]]
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* [[1983 AIME Problems|Back to Exam]]
  
 
== See also ==
 
== See also ==
−
* [[1983 AIME Problems]]
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 00:13, 24 July 2006

Problem

The numbers $1447$, $1005$, and $1231$ have something in common. Each is a four-digit number beginning with $1$ 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,

$11xy,\qquad 1x1y,\qquad1xy1.$

Because the number must have exactly two identical digits, $x\neq y$, $x\neq1$, and $y\neq1$. Hence, there are $3\cdot9\cdot8=216$ numbers of this form.

Suppose the two identical digits are not one. Therefore, consider the following possibilities,

$1xxy,\qquad1xyx,\qquad1yxx.$

Again, $x\neq y$, $x\neq 1$, and $y\neq 1$. There are $3\cdot9\cdot8=216$ numbers of this form too.

Thus, the desired answer is $216+216=432$.

See also

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