1988 AIME Problems/Problem 6

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Problem

It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*).

Solution

Solution 1 (specific)

Let the coordinates of the square at the bottom left be $(0,0)$ , the square to the right $(1,0)$ , etc.

Label the leftmost column (from bottom to top) $0, a, 2a, 3a, 4a$ and the bottom-most row (from left to right) $0, b, 2b, 3b, 4b$ . Our method will be to use the given numbers to set up equations to solve for $a$ and $b$ , and then calculate $(*)$ .

$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & * & \\\hline 3a & 74 & & & \\\hline 2a & &...$

We can compute the squares at the intersections of two existing numbers in terms of $a$ and $b$ ; two such equations will give us the values of $a$ and $b$ . On the fourth row from the bottom, the common difference is $74 - 3a$ , so the square at $(2,3)$ has a value of $148 - 3a$ . On the third column from the left, the common difference is $103 - 2b$ , so that square also has a value of $2b + 3(103 - 2b) = 309 - 4b$ . Equating, we get $148 - 3a = 309 - 4b \Longrightarrow 4b - 3a = 161$ .

Now we compute the square $(2,2)$ . By rows, this value is simply the average of $2a$ and $186$ , so it is equal to $\frac{2a + 186}{2} = a + 93$ . By columns, the common difference is $103 - 2b$ , so our value is $206 - 2b$ . Equating, $a + 93 = 206 - 2b \Longrightarrow a + 2b = 113$ .

Solving $\begin{eqnarray*}4b - 3a &=& 161\\a + 2b &=& 113$

gives $a = 13$ , $b = 50$ . Now it is simple to calculate $(4,3)$ . One way to do it is to see that $(2,2)$ has $206 - 2c = 106$ and $(4,2)$ has $186$ , so $(3,2)$ has $\frac{106 + 186}{2} = 146$ . Now, $(3,0)$ has $3b = 150$ , so $(3,2) = \frac{(3,0) + (3,4)}{2} = \Longrightarrow (3,4) = * = 142$ .

Solution 2 (general)

First, let $a =$ the number to be placed in the first column, fourth row. Let $b =$ the number to be placed in the second column, fifth row. We can determine the entire first column and fifth row in terms of $a$ and $b$ :

$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & & \\\hline 3a & & & & \\\hline 2a & & ...$

Next, let $a + b + c =$ the number to be placed in the second column, fourth row. We can determine the entire second column and fourth row in terms of $a$ , $b$ , and $c$ :

$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & & & \\\hline 3a & 3a + b + 3c & & & ...$

We have now determined at least two values in each row and column. We can finish the table without introducing any more variables:

$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & 4a + 2b + 8c & 4a + 3b + 12c & 4a + 4b + 16c \\\hlin...$

We now have a system of equations.

$3a + b + 3c = 74$

$2a + 4b + 8c = 186$

$a + 2b + 2c = 103$

Solving, we find that $(a,b,c) = (13,50, - 5)$ . The number in the square marked by the asterisk is $4a + 3b + 12c = \boxed{142}$

1988 AIME (Problems • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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