Difference between revisions of "2017 AIME I Problems/Problem 11"

Revision as of 02:14, 5 March 2018

Problem 11

Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ , $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ , $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$ .

Solution 1

We know that if $5$ is a median, then $5$ will be the median of the medians.

WLOG, assume $5$ is in the upper left corner. One of the two other values in the top row needs to be below $5$ , and the other needs to be above $5$ . This can be done in $4\cdot4\cdot2=32$ ways. The other $6$ can be arranged in $6!=720$ ways. Finally, accounting for when $5$ is in every other space, our answer is $32\cdot720\cdot9$ , which is $207360$ . But we only need the last $3$ digits, so $\boxed{360}$ 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 $m=4$

There is a $\frac{15}{28}$ chance that exactly one of 1, 2, 3 is in the same row.

There is a $\frac{2}{5}$ chance that the other two smaller numbers end up in the same row.

$9!(1-2*\frac{15}{28}*\frac{2}{5})=362880*\frac{4}{7}=207\boxed{360}$ .

Solution 3

We will make sure to multiply by $3!$ in the end to account for all the possible permutation of the rows.

WLOG, let $5$ be present in the Row # $1$ .

Notice that $5$ MUST be placed with a number lower than it and a number higher than it.

This happens in $4\cdot4$ ways. You can permutate Row # $1$ in $3!$ ways.

Now, take a look at Row $2$ and Row $3$ .

Because there are $6$ numbers to choose from now, you can assign #'s to Row's #2&3 in

$\frac{\binom{6}{3}\cdot\binom{3}{3}}{2}$ ways. There are $3!\cdot3!$ ways to permutate the numbers in the individual Rows.

Hence, our answer is $3!(4\cdot4\cdot3!\cdot{10}\cdot{3!}\cdot{3!})=3!(16\cdot60\cdot36)=3!(34560)\implies{207\boxed{360}}$

@@ Line 8: / Line 8: @@
 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.
-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.
-~Solution by SuperSaiyanOver9000, mathics42
+~Solution by SuperSaiyanOver9000, mathics42, Kinglogic
 ==Solution 2==

2017 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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