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Problem of the Week
Problem D and Solution
What’s in That Square?

Problem

Fourteen squares are placed in a row forming the grid below. Each square is to be filled with a positive integer, according to the following rules.

  1. The product of any four integers in adjacent squares is \(120\).

  2. Integers may appear more than once in the grid.

Four of the squares are already filled with a positive integer, as shown. Determine all possible values of \(x\).

The third square has a 2, the sixth square has a 4, the
ninth square has the variable x, and the twelfth square has a 3. The
remaining squares are empty.

Solution

In both solutions, let \(a_1\) be the positive integer in the first square, \(a_2\) the positive integer in the second square, \(a_3\) be the positive integer in the third square, \(a_4\) the positive integer in the fourth square, and so on.

Solution 1

Consider squares \(3\) to \(6\). Since the product of any four adjacent integers is \(120\), we have \(2\times a_4\times a_5 \times 4 = 120\). Therefore, \(a_4\times a_5 = \frac{120}{2\times 4} = 15\). Since \(a_4\) and \(a_5\) are positive integers, there are four possibilities: \(a_4 = 1\) and \(a_5 = 15\), or \(a_4 = 15\) and \(a_5 = 1\), or \(a_4 = 3\) and \(a_5 = 5\), or \(a_4 = 5\) and \(a_5 = 3\).

In each of the four cases, we will have \(a_7 = 2\). We can see why by considering squares \(4\) to \(7\). We have \(a_4\times a_5 \times 4 \times a_7 = 120\), or \(15 \times 4 \times a_7 = 120\), since \(a_4\times a_5 = 15\). Therefore, \(a_7 = \frac{120}{15\times 4} = 2\).

Therefore, the only possible value for \(x\) is \(x=5\).

Solution 2

You may have noticed a pattern for the \(a_i\)’s in Solution \(1\). We will explore this pattern.

Since the product of any four adjacent integers is \(120\), \(a_1a_2a_3a_4 = a_2a_3a_4a_5 = 120\). Since both sides are divisible by \(a_2a_3a_4\), and each is a positive integer, then \(a_1 = a_5\).
Similarly, \(a_2a_3a_4a_5 = a_3a_4a_5a_6 = 120\), and so \(a_2 = a_6\).
In general, \(a_na_{n+1}a_{n+2}a_{n+3} = a_{n+1}a_{n+2}a_{n+3}a_{n+4}\), and so \(a_n = a_{n+4}\).
We can use this along with the given information to fill out the entries in the squares as follows:

The entries in the fourteen squares, in order, are x, 4, 2,
3, x, 4, 2, 3, x, 4, 2, 3, x, 4.

Therefore, \(4\times 2\times 3\times x = 120\) and so \(x = \frac{120}{4\times 2\times 3} = 5\).