Difference between revisions of "1983 AIME Problems/Problem 10"
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− | One of the three numbers must be <math>1</math>, and the other two numbers must be the same number, but cannot be <math>1</math> (That will be dealt with in case 4). There are <math>3</math> choices to put the <math>1</math>, and there are <math>9</math> choices ( | + | One of the three numbers must be <math>1</math>, and the other two numbers must be the same number, but cannot be <math>1</math> (That will be dealt with in case 4). There are <math>3</math> choices to put the <math>1</math>, and there are <math>9</math> choices (any number except for <math>1</math>) to pick the other number that is repeated, so a total of <math>3\cdot9=27</math> numbers. |
Case 3: | Case 3: |
Latest revision as of 15:03, 13 December 2020
Contents
Problem
The numbers , and have something in common: each is a -digit number beginning with that has exactly two identical digits. How many such numbers are there?
Solution
Solution 1
Suppose that the two identical digits are both . Since the thousands digit must be , only one of the other three digits can be . This means the possible forms for the number are
Because the number must have exactly two identical digits, , , and . Hence, there are numbers of this form.
Now suppose that the two identical digits are not . Reasoning similarly to before, we have the following possibilities:
Again, , , and . There are numbers of this form.
Thus the answer is .
Solution 2
Consider a sequence of digits instead of a -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 . This means we can find all possible sequences with one digit repeated twice, and then divide by .
If we let the three distinct digits of the sequence be and , with repeated twice, we can make a table with all possible sequences:
There are possible sequences.
Next, we can see how many ways we can pick , , and . This is , because there are digits, from which we need to choose with regard to order. This means there are sequences of length with one digit repeated. We divide by 10 to get as our answer.
Solution 3 (Complementary Counting)
We'll use complementary counting. We will split up into cases: (1) no number is repeated, (2) numbers are repeated, and other numbers are repeated, (3) numbers are repeated, or (4) numbers are repeated.
Case 1: There are choices for the hundreds digit (it cannot be ), choices for the tens digit (it cannot be or what was chosen for the hundreds digit), and for the units digit. This is a total of numbers.
Case 2: One of the three numbers must be , and the other two numbers must be the same number, but cannot be (That will be dealt with in case 4). There are choices to put the , and there are choices (any number except for ) to pick the other number that is repeated, so a total of numbers.
Case 3: We will split it into subcases: one where is repeated times, and one where another number is repeated times. When is repeated times, then one of the digits is not . There are choices for that number, and choices for its location,so a total of numbers. When a number other than is repeated times, then there are choices for the number, and you don't have any choices on where to put that number. So in Case 3 there are numbers
Case 4: There is only number: .
There are a total of -digit numbers that begin with (from to ), so by complementary counting you get numbers.
Solution by Kinglogic
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 |