Difference between revisions of "2017 AIME I Problems/Problem 11"
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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. | 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. | ||
| − | ~Solution by SuperSaiyanOver9000, mathics42 | + | ~Solution by SuperSaiyanOver9000, mathics42 |
==Solution 2== | ==Solution 2== | ||
Revision as of 13:11, 8 October 2018
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
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 permute 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. 
