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Difference between revisions of "2002 Indonesia MO Problems/Problem 5"

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[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 23:09, 3 August 2018

Problem

Nine of the following ten numbers: $4,5,6,7,8,12,13,16,18,19$ are going to be filled into empty spaces in the $3 \times 5$ table shown below. After all spaces are filled, the sum of the numbers on each row will be the same. And so with the sum of the numbers on each column, will also be the same. Determine all possible fillings.

$\begin{array} {|c|c|c|} \cline{1-3} 10 & & \\ \cline{1-3} & & 9 \\ \cline{1-3} & 3 & \\ \cline{1-3} 11 & & 17 \\ \cline{1-3} & 20 & \\ \cline{1-3} \end{array}$

Solution

The sum of the numbers currently in the grid and all ten numbers is $3+4+ \cdots 13 + 16 + 17 \cdots 20 = \tfrac{16 \cdot 11}{2} + \tfrac{36 \cdot 5}{2} = 178$. Note that the sum of the numbers in each row and the sum of the numbers in each column are whole numbers, so the sum of all the numbers that would be in the grid is a multiple of $3$ and $5$. Because $178 \equiv 3 \pmod{5}$ and $177 \equiv 1 \pmod{3}$, we know that a number not in the grid is congruent to $3$ modulo $5$ and $1$ modulo $3$. The only number of the ten numbers that meets the criteria is $13$, so $13$ will not be in the grid.

That makes the sum of the numbers in the grid $165$, so the sum of each row is $33$ and the sum of each column is $55$. Thus, $5$ must appear between $11$ and $17$. Also, the remaining two numbers in the middle column sum to $27$, The only two numbers from the remaining list that add up to $27$ are $19$ and $8$.

If $8$ is on the top row of the middle column, then the right column of the top row must have $15$, which couldn't happen. Therefore, $19$ is on the top row with $4$ to the right of it, and $8$ is on the second row of the middle column with $16$ to the left of it.

The remaining numbers are $6$, $7$, $12$, and $18$. The remaining numbers of the last row must sum to $13$, so $6$ and $7$ are in the bottom row while $12$ and $18$ are on the third row. Also, the remaining numbers of the last column must sum to $25$, so $18$ and $7$ are on the last column.

With this in mind, there is only one possible configuration that satisfies the conditions (shown below).

$\begin{array} {|c|c|c|} \cline{1-3} 10 & 19 & 4 \\ \cline{1-3} 16 & 8 & 9 \\ \cline{1-3} 12 & 3 & 18 \\ \cline{1-3} 11 & 5 & 17 \\ \cline{1-3} 6 & 20 & 7 \\ \cline{1-3} \end{array}$

See Also

2002 Indonesia MO (Problems)
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 • 7 Followed by
Problem 6
All Indonesia MO Problems and Solutions
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