Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 13"
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Consider <math>2005</math> people, who will be separated into group <math>1</math>, group <math>2</math>, and group <math>3</math>. Furthermore, one person in group <math>1</math> will be cool, and one person will be smart. (They may be the same people). | Consider <math>2005</math> people, who will be separated into group <math>1</math>, group <math>2</math>, and group <math>3</math>. Furthermore, one person in group <math>1</math> will be cool, and one person will be smart. (They may be the same people). | ||
− | Consider only putting <math>k</math> people into group <math>1</math>. There are <math>\dbinom{2005}{k}</math> ways this can be done. For the remaining <math>2005-k</math> people, there are one of two groups they can be in, namely group <math>2</math> and group <math>3</math>. This means that there are <math>2^{2005-k}</math> ways this can be done. There are <math>k</math> ways to determine who is cool, and <math>k</math> ways to determine who is | + | Consider only putting <math>k</math> people into group <math>1</math>. There are <math>\dbinom{2005}{k}</math> ways this can be done. For the remaining <math>2005-k</math> people, there are one of two groups they can be in, namely group <math>2</math> and group <math>3</math>. This means that there are <math>2^{2005-k}</math> ways this can be done. There are <math>k</math> ways to determine who is cool, and <math>k</math> ways to determine who is smart. This is <math>k^2</math>. As <math>k</math> ranges from <math>1</math> to <math>2005</math>, we will get all such scenarios. This means that the number of ways that this can be done is also <math>3^{2005}S</math>. |
Another way to count this is two split it up into two cases. | Another way to count this is two split it up into two cases. |
Latest revision as of 13:53, 19 July 2020
Contents
Problem
Let denote the value of the sum
Determine the remainder obtained when is divided by .
Solution 1
Let . Let . Then note that , so taking the derivative and multiplying by gives . Taking the derivative and multiplying by again gives . Now note that . Then we get , so , so .
Solution 2
Let the wanted sum be . We will simplify the expression into: . A counting argument will be provided to compute this.
Consider people, who will be separated into group , group , and group . Furthermore, one person in group will be cool, and one person will be smart. (They may be the same people).
Consider only putting people into group . There are ways this can be done. For the remaining people, there are one of two groups they can be in, namely group and group . This means that there are ways this can be done. There are ways to determine who is cool, and ways to determine who is smart. This is . As ranges from to , we will get all such scenarios. This means that the number of ways that this can be done is also .
Another way to count this is two split it up into two cases.
Case 1: person is both cool and smart. There are ways to choose this person. The remaining people have a choice of one of groups, making ways.
Case 2: person is cool, and another person in smart. There are ways to choose who is cool, and ways to choose who is smart. The remaining people have a choice of one of groups, making .
Thus, we have:
Thus, the answer is
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |