1983 AIME Problems/Problem 10

Revision as of 21:25, 19 July 2018 by Ultranitedu (talk | contribs) (Solution 2)

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

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.

See Also

1983 AIME (ProblemsAnswer KeyResources)
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