2017 AIME I Problems/Problem 11
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
. 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
\cdot
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 #$2&3$ (Error compiling LaTeX. Unknown error_msg) 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 |
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