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

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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?
 
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?
  
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== Solution ==
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== Solution 1==
 
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,
  
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Thus, the desired answer is <math>216+216=\boxed{432}</math>.
 
Thus, the desired answer is <math>216+216=\boxed{432}</math>.
  
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== Solution 2 ==
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Consider a sequence of four digits instead of a four digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is <math>\frac{1}{10}</math>. This means we can find all possible sequences with one digit repeated twice, and then divide by <math>10</math>.
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If we let the three distinct digits of the sequence be <math>a, b,</math> and <math>c,</math> with <math>a</math> repeated twice, we can make a table with all possible sequences:
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<cmath>\begin{tabular}{ccc}
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aabc & abac & abca \\
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baac & baca & \\
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bcaa && \\
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\end{tabular} </cmath>
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There are <math>6</math> possible sequences.
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Next, we can see how many ways we can pick <math>a, b,</math> and <math>c</math>. This is <math>10(9)(8) = 720</math> because there are <math>10</math> digits, and we need to choose <math>3</math> with regard to order. This means there are <math>720(9) = 4320</math> sequences of length <math>4</math> with one digit repeated. We divide by 10 to get <math>\boxed{432}</math> as our answer.
 
== See Also ==
 
== See Also ==
 
{{AIME box|year=1983|num-b=9|num-a=11}}
 
{{AIME box|year=1983|num-b=9|num-a=11}}

Revision as of 01:06, 26 July 2017

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 1

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.

Thus, the desired answer is $216+216=\boxed{432}$.

Solution 2

Consider a sequence of four digits instead of a four digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is $\frac{1}{10}$. This means we can find all possible sequences with one digit repeated twice, and then divide by $10$.

If we let the three distinct digits of the sequence be $a, b,$ and $c,$ with $a$ repeated twice, we can make a table with all possible sequences:

\[\begin{tabular}{ccc} aabc & abac & abca \\ baac & baca & \\ bcaa && \\ \end{tabular}\]

There are $6$ possible sequences.

Next, we can see how many ways we can pick $a, b,$ and $c$. This is $10(9)(8) = 720$ because there are $10$ digits, and we need to choose $3$ with regard to order. This means there are $720(9) = 4320$ sequences of length $4$ with one digit repeated. We divide by 10 to get $\boxed{432}$ as our answer.

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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