Difference between revisions of "2017 AIME I Problems/Problem 11"
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| + | ==Problem 11== | ||
| + | Consider arrangements of the <math>9</math> numbers <math>1, 2, 3, \dots, 9</math> in a <math>3 \times 3</math> array. For each such arrangement, let <math>a_1</math>, <math>a_2</math>, and <math>a_3</math> be the medians of the numbers in rows <math>1</math>, <math>2</math>, and <math>3</math> respectively, and let <math>m</math> be the median of <math>\{a_1, a_2, a_3\}</math>. Let <math>Q</math> be the number of arrangements for which <math>m = 5</math>. Find the remainder when <math>Q</math> is divided by <math>1000</math>. | ||
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| + | ==Solution 1== | ||
| + | |||
We know that if <math>5</math> is a median, then <math>5</math> will be the median of the medians. | We know that if <math>5</math> is a median, then <math>5</math> will be the median of the medians. | ||
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| − | Solution 2 (Complementary Counting with probability) | + | ==Solution 2== |
| + | (Complementary Counting with probability) | ||
Notice that m can only equal 4, 5, or 6, and 4 and 6 are symmetric. | Notice that m can only equal 4, 5, or 6, and 4 and 6 are symmetric. | ||
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<math>9!(1-2*\frac{15}{28}*\frac{2}{5})=362880*\frac{4}{7}=207\boxed{360}</math>. | <math>9!(1-2*\frac{15}{28}*\frac{2}{5})=362880*\frac{4}{7}=207\boxed{360}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AIME box|year=2017|n=I|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Revision as of 19:26, 8 March 2017
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 
. 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.
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
