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

Revision as of 19:24, 24 December 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 have 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.

@@ Line 10: / Line 10: @@
 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.
-Now suppose that the two identical digits are not <math>1</math>. Reasoning similarly to before, we the following possibilities:
+Now suppose that the two identical digits are not <math>1</math>. Reasoning similarly to before, we have the following possibilities:
 <div style="text-align:center;"><math>1xxy,\qquad1xyx,\qquad1yxx.</math></div>

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
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