Difference between revisions of "1983 AIME Problems/Problem 10"
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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 four-digit number beginning with <math>1</math> that has exactly two identical digits. How many such numbers are there? | ||
− | == Solution == | + | == 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 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, | ||
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Thus, the desired answer is <math>216+216=\boxed{432}</math>. | Thus, the desired answer is <math>216+216=\boxed{432}</math>. | ||
+ | == 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>. | ||
+ | |||
+ | 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} | ||
+ | aabc & abac & abca \\ | ||
+ | baac & baca & \\ | ||
+ | bcaa && \\ | ||
+ | \end{tabular} </cmath> | ||
+ | |||
+ | 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(9) = 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 == | ||
{{AIME box|year=1983|num-b=9|num-a=11}} | {{AIME box|year=1983|num-b=9|num-a=11}} |
Revision as of 02:06, 26 July 2017
Contents
Problem
The numbers ,
, and
have something in common. Each is a four-digit number beginning with
that has exactly two identical digits. How many such numbers are there?
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,
![$11xy,\qquad 1x1y,\qquad1xy1$](http://latex.artofproblemsolving.com/6/1/8/618de0dca3122f7ae4c49bd44b021966d9723848.png)
Because the number must have exactly two identical digits, ,
, and
. Hence, there are
numbers of this form.
Suppose the two identical digits are not one. Therefore, consider the following possibilities,
![$1xxy,\qquad1xyx,\qquad1yxx.$](http://latex.artofproblemsolving.com/2/c/c/2ccebb04bbfa82459be3147685b2846acfc824d8.png)
Again, ,
, and
. There are
numbers of this form.
Thus, the desired answer is .
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 . 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, and 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.
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 |