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

Revision as of 18:41, 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 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 $720(6) = 4320$ sequences of length $4$ with one digit repeated. We divide by 10 to get $\boxed{432}$ as our answer.

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

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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