Difference between revisions of "1988 AIME Problems/Problem 6"

Latest revision as of 22:06, 20 November 2023

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 (*).

Solutions

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 $(*)$ .

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

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

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 $206 - 2b = 106$ and $(4,2)$ has $186$ , so $(3,2)$ has $\frac{106 + 186}{2} = 146$ . Now, $(3,0)$ has $3b = 150$ , so $(3,2) = \frac{(3,0) + (3,4)}{2} \Longrightarrow (3,4) = * = \boxed{142}$ .

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

Solution 3 (Only one variable)

We begin with the table that was given to us and add in the following arithmetic progression on the bottom:

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

Since all the rows and columns satisfy an arithmetic progression, we have the following:

$\begin{tabular}[b]{|c|c|c|c|c|}\hline & & 412 - 6x & 392 - 5x & 372 - 4x \\ \hline & 74 & 309 - 4x & 294 - 3x & 279 - 2x \\ \hline & & 206 - 2x & 196 - x & 186 \\ \hline & & 103 & 98 + x & 2x + 93 \\ \hline 0 & x & 2x & 3x & 4x \\ \hline \end{tabular}$

We can solve for $x$ in the 2nd row, namely $324 - 5x = 74$ because the arithmetic progression from left to right has difference $x - 15$ . Therefore, we have $x = 50$ , and because the desired asterisk is $392 - 5x$ , the answer is $392 - 250$ = $\boxed {142}$ .

~Arcticturn

@@ Line 68: / Line 68: @@
 ==Solution 3 (Only one variable)==
-We begin with the table that was given to us
+We begin with the table that was given to us and add in the following arithmetic progression on the bottom:
 <math>\begin{tabular}[b]{|c|c|c|c|c|}\hline & & & * & \\
@@ Line 85: / Line 85: @@
 \hline 0 & x & 2x & 3x & 4x \\
 \hline \end{tabular}</math>
+We can solve for <math>x</math> in the 2nd row, namely <math>324 - 5x = 74</math> because the arithmetic progression from left to right has difference <math>x - 15</math>. Therefore, we have <math>x = 50</math>, and because the desired asterisk is <math>392 - 5x</math>, the answer is <math>392 - 250</math> = <math>\boxed {142}</math>.
+~Arcticturn
 == See also ==
 {{AIME box|year=1988|num-b=5|num-a=7}}
 {{MAA Notice}}

1988 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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