Difference between revisions of "2017 AIME I Problems/Problem 11"
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Revision as of 10:18, 7 June 2022
Contents
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 
1. There is a 
chance that exactly one of 1, 2, 3 is in the same row with 4.
There are 3 ways to select which of the smaller numbers will get in the row, and then 5
ways to select the number larger than 4.
2. There is a 
chance that the other two smaller numbers end up in the same row.
There are 2 ways to select the row that the two smaller number are in, and then 
ways
to place the smaller numbers in the 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 
Solution 4
We see that if one of the medians is 5, then there are two remaining numbers greater than 5 and two less than 5, so it follows that 
. There are 3 ways to choose which row to have 5 in, 
ways to choose the other two numbers in that row, 
ways to arrange the numbers in that row, and ways for the remaining numbers, for our answer is 
.
-Stormersyle
Solution 5
We take the grid, and we do a bunch of stuff with it. First, we sort each row, smallest on the left, largest on the right, then we arrange these 3 rows such that the middle #s are increasing, from top to bottom. Thus, we get that the cell in the very center of the grid must be 5. (We need to multiply by 1296 at the end)
Let S mean a number that is < 5, L mean a number > 5.
In the 2 blank corners, one can be S, and the other one has to be L.
If the top right corner is S, there are 16 ways, otherwise, there are 144 ways. (This is left as an exercise to the reader).
Thus, there are 160 * 1296 total configurations, which gets us an answer of 
-AlexLikeMath
Solution 6
We note that if is a median of one of the rows, than . First, focus on the row with in it. There are 
ways to choose the other numbers in that row and then ways to order it. Now, clearly, there are 
ways to put the other numbers into the remaining slots so 
. Hence, our answer is 
.
(Solution by pleaseletmetwin, but not added to the wiki by pleaseletmetwin)
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. 
