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

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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 $32\cdot720\cdot9$ . But we only need the last $3$ digits, so $\boxed{360}$ 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 $m=4$

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 $4$ \cdot $4$ 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 #$2&3$ (Error compiling LaTeX. Unknown error_msg) 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 25: / Line 25: @@
 <math>9!(1-2*\frac{15}{28}*\frac{2}{5})=362880*\frac{4}{7}=207\boxed{360}</math>.
+==Solution 3==
+We will make sure to multiply by <math>3!</math> in the end to account for all the possible permutation of the rows.
+WLOG, let <math>5</math> be present in the Row #<math>1</math>.
+Notice that <math>5</math> MUST be placed with a number lower than it and a number higher than it.
+This happens in <math>4</math>\cdot<math>4</math> ways. You can permutate Row #<math>1</math> in <math>3!</math> ways.
+Now, take a look at Row <math>2</math> and Row <math>3</math>.
+Because there are <math>6</math> numbers to choose from now, you can assign #'s to Row's #<math>2&3</math> in <math>\frac{\binom{6}{3}\cdot\binom{3}{3}}{2}</math> ways. There are <math>3!\cdot3!</math> ways to permutate the numbers in the individual Rows.
+Hence, our answer is <math>3!(4\cdot4\cdot3!\cdot{10}\cdot{3!}\cdot{3!}=3!(16\cdot60\cdot36)=3!(34560)\implies{207\boxed{360}}</math>
 ==See Also==
 {{AIME box|year=2017|n=I|num-b=10|num-a=12}}
 {{MAA Notice}}

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