Difference between revisions of "2017 AIME I Problems/Problem 11"
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WLOG, assume <math>5</math> is in the upper left corner. One of the two other values in the top row needs to be below <math>5</math>, and the other needs to be above <math>5</math>. This can be done in <math>4\cdot4\cdot2=32</math> ways. | WLOG, assume <math>5</math> is in the upper left corner. One of the two other values in the top row needs to be below <math>5</math>, and the other needs to be above <math>5</math>. This can be done in <math>4\cdot4\cdot2=32</math> ways. | ||
The other <math>6</math> can be arranged in <math>6!=720</math> ways. | The other <math>6</math> can be arranged in <math>6!=720</math> ways. | ||
− | Finally, accounting for when <math>5</math> is in every other space, our answer is <math>32\cdot720\cdot9</math>. But we only need the last <math>3</math> digits, so <math>\boxed{360}</math> is our answer. | + | Finally, accounting for when <math>5</math> is in every other space, our answer is <math>32\cdot720\cdot9</math>, which is <math>207360</math>. But we only need the last <math>3</math> digits, so <math>\boxed{360}</math> is our answer. |
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+ | ~Solution by SuperSaiyanOver9000, mathics42, Kinglogic | ||
==Solution 2== | ==Solution 2== |
Revision as of 02:14, 5 March 2018
Contents
[hide]Problem 11
Consider arrangements of the numbers in a array. For each such arrangement, let , , and be the medians of the numbers in rows , , and respectively, and let be the median of . Let be the number of arrangements for which . Find the remainder when is divided by .
Solution 1
We know that if is a median, then will be the median of the medians.
WLOG, assume is in the upper left corner. One of the two other values in the top row needs to be below , and the other needs to be above . This can be done in ways. The other can be arranged in ways. Finally, accounting for when is in every other space, our answer is , which is . But we only need the last digits, so is our answer.
~Solution by SuperSaiyanOver9000, mathics42, Kinglogic
Solution 2
(Complementary Counting with probability)
Notice that m can only equal 4, 5, or 6, and 4 and 6 are symmetric.
WLOG let
There is a chance that exactly one of 1, 2, 3 is in the same row.
There is a chance that the other two smaller numbers end up in the same row.
.
Solution 3
We will make sure to multiply by in the end to account for all the possible permutation of the rows.
WLOG, let be present in the Row #.
Notice that MUST be placed with a number lower than it and a number higher than it.
This happens in ways. You can permutate Row # in ways.
Now, take a look at Row and Row .
Because there are numbers to choose from now, you can assign #'s to Row's #2&3 in
ways. There are ways to permutate the numbers in the individual Rows.
Hence, our answer is
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.