Difference between revisions of "2009 AIME I Problems/Problem 8"
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− | When computing <math>N</math>, the number <math>2^x</math> will be added <math>x</math> times (for terms <math>2^x-2^0</math>, <math>2^x-2^1</math>, ..., <math>2^x - 2^{x-1}</math>), and subtracted <math>10-x</math> times. Hence <math>N</math> can be computed as <math>N=10\cdot 2^{10} + 8\cdot 2^9 + 6\cdot 2^8 + \cdots - 8\cdot 2^1 - 10\cdot 2^0</math>. | + | When computing <math>N</math>, the number <math>2^x</math> will be added <math>x</math> times (for terms <math>2^x-2^0</math>, <math>2^x-2^1</math>, ..., <math>2^x - 2^{x-1}</math>), and subtracted <math>10-x</math> times. Hence <math>N</math> can be computed as <math>N=10\cdot 2^{10} + 8\cdot 2^9 + 6\cdot 2^8 + \cdots - 8\cdot 2^1 - 10\cdot 2^0</math>. Evaluating <math>N\bmod 1000</math> yields: |
<cmath> | <cmath> |
Revision as of 09:31, 23 January 2023
Contents
[hide]Problem
Let . Consider all possible positive differences of pairs of elements of . Let be the sum of all of these differences. Find the remainder when is divided by .
Solution 1 (Extreme bash)
Find the positive differences in all pairs and you will get .
Solution 2
When computing , the number will be added times (for terms , , ..., ), and subtracted times. Hence can be computed as . Evaluating yields:
Solution 3
This solution is a generalized approach that work when is replaced by other values.
In similarity to the logic in Solution 2, .
Let and let . Then .
can be calculated easily by geometric series with sum . Hence .
We can compute using a trick known as the change of summation order.
Imagine writing down a table that has rows with labels 0 to 10. In row , write the number into the first columns. You will get a triangular table. Obviously, the row sums of this table are of the form , and therefore the sum of all the numbers is precisely .
Now consider the ten columns in this table. Let's label them 1 to 10. In column , you have the values to , each of them once. And this is just a geometric series with the sum . We can now sum these column sums to get . Hence we have . This simplifies to .
Hence .
Then .
Solution 4
Consider the unique differences . Simple casework yields a sum of . This method generalizes nicely as well.
Video Solution
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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