1988 AIME Problems/Problem 6

Revision as of 20:18, 26 September 2007 by Azjps (talk | contribs) (Solution 1 (specific): whoops)

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 (*).

1988 AIME-6.png

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 using the given numbers to set up equations to solve for $a$ and $b$, and then calculate $(*)$.

$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & * & \\ \hline 3a & 74 & & & \\ \hline 2a & & & & 186 \\ \hline a & & 103 & & \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$

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 (Error compiling LaTeX. ! LaTeX Error: \begin{eqnarray*} on input line 20 ended by \end{document}.)

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 $b$:

$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & & \\ \hline 3a & & & & \\ \hline 2a & & & & \\ \hline a & & & & \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$

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 $c$:

$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & & & \\ \hline 3a & 3a + b + 3c & & & \\ \hline 2a & 2a + b + 2c & & & \\ \hline a & a + b + c & a + 2b + 2c & a + 3b + 3c & a + 4b + 4c \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$

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

$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & 4a + 2b + 8c & 4a + 3b + 12c & 4a + 4b + 16c \\ \hline 3a & 3a + b + 3c & 3a + 2b + 6c & 3a + 3b + 9c & 3a + 4b + 12c \\ \hline 2a & 2a + b + 2c & 2a + 2b + 4c & 2a + 3b + 6c & 2a + 4b + 8c \\ \hline a & a + b + c & a + 2b + 2c & a + 3b + 3c & a + 4b + 4c \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$

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 $4a + 3b + 12c = \boxed{142}$

See also

1988 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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