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

m (< center > -> < div >)
m (→‎Solution)
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<div style="text-align:center;"><math>11xy,\qquad 1x1y,\qquad1xy1.</math></div>
 
<div style="text-align:center;"><math>11xy,\qquad 1x1y,\qquad1xy1.</math></div>
  
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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.
+
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.
  
 
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,
Line 13: Line 13:
 
<div style="text-align:center;"><math>1xxy,\qquad1xyx,\qquad1yxx.</math></div>
 
<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 as well.
  
 
Thus, the desired answer is <math>216+216=432</math>.
 
Thus, the desired answer is <math>216+216=432</math>.

Revision as of 17:42, 1 March 2009

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 as well.

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

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
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