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1983 AIME Problems/Problem 10: Difference between revisions

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==Solution==
==Solution==
===Solution 1===
===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 that the two identical digits are both <math>1</math>. Since the thousands digit must be <math>1</math>, only one of the other three digits can be <math>1</math>. This means the possible forms for the number are


<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>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>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,
Now suppose that the two identical digits are not <math>1</math>. Reasoning similarly to before, we the following possibilities:


<div style="text-align:center;"><math>1xxy,\qquad1xyx,\qquad1yxx.</math></div>
<div style="text-align:center;"><math>1xxy,\qquad1xyx,\qquad1yxx.</math></div>
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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.
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.


Thus, the desired answer is <math>216+216=\boxed{432}</math>.
Thus the answer is <math>216+216=\boxed{432}</math>.


===Solution 2===  
===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 <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>.  
Consider a sequence of <math>4</math> digits instead of a <math>4</math>-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>.  


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:
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:


<cmath>\begin{tabular}{ccc}
<cmath>\begin{tabular}{ccc}
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There are <math>6</math> possible sequences.  
There are <math>6</math> possible sequences.  


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(6) = 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.
Next, we can see how many ways we can pick <math>a</math>, <math>b</math>, and <math>c</math>. This is <math>10(9)(8) = 720</math>, because there are <math>10</math> digits, from which we need to choose <math>3</math> with regard to order. This means there are <math>720(6) = 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 ==

Revision as of 18:44, 15 February 2019

Problem

The numbers $1447$, $1005$ and $1231$ have something in common: each is a $4$-digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

Solution

Solution 1

Suppose that the two identical digits are both $1$. Since the thousands digit must be $1$, only one of the other three digits can be $1$. This means the possible forms for the number are

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

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.

Now suppose that the two identical digits are not $1$. Reasoning similarly to before, we 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 answer is $216+216=\boxed{432}$.

Solution 2

Consider a sequence of $4$ digits instead of a $4$-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, from which we need to choose $3$ with regard to order. This means there are $720(6) = 4320$ 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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