Football Fanfare
The students in the pep club are designing a "card stunt" for the halftime show for the next
football game. They decide to use 108 square cards to fill in the area of the rectangular
display. Each card will contain part of a picture or message. Each volunteer will hold up one of
the 108 cards so the group will form complete pictures.
Here are two examples of how some smaller card stunts might look:
Question: Which arrangement would you choose for displaying a message? What are all
the possible seating arrangements (how many rows and how many seats in each row) for
the 108 students? Use the space below to explain your response.
Extra: For the last game of the season, they decide to make a much larger display by enlisting
the help of parents, siblings and friends. They get 500 people to agree to help with the stunt.
First, guess how many rectangular arrangements are possible and tell us why that was your
guess. Next, figure out the possible arrangements. Were you surprised by the results? Explain
your thinking.
Expected Solution
The possible seating arrangements are: 1 row with 108 seats, 2 rows with 54 seats, 3 rows
with 36 seats, 4 rows with 27 seats, 6 rows with 18 seats, 9 rows with 12 seats, 108 rows with
1 seat, 54 rows with 2 seats, 36 rows with 3 seats, 27 rows with 4 seats, 18 rows with 6 seats,
and 12 rows with 9 seats.
The arrangement I would use for displaying a message would be 6 by 18.
To get the answer for the first problem, I knew that all I needed to do was find all of the factor
pairs of 108. To do that, I went through the numbers one to twelve with their divisibility rules
to see if they were factors of 108 (other than 7 because 7's rule is more complicated than the
division itself). The numbers that 108 could be divided by evenly were 1, 2, 3, 4, 6, 9, and 12.
Since 9 and 12 were a factor pair by themselves, I knew I had to look no further to find at least
half of the pairs because 9 and 12 are in the middle of the factor list.
Then I divided 108 by the rest of the factors to find the other factor of the pair. The factor
pairs I found were: 1 x 108 , 2 x 54, 3 x 36, 4 x 27, 6 x 18, and 9 x 12, which I had already found
before. Then I added 108 x 1, 54 x 2, 36 x 3, 27 x 4, 18 x 6, and 12 x 9 on to the list because
those would make different rectangles or seating arrangements. So, the final seating
arrangements the pep club could use are 1 row with 108 seats, 2 rows with 54 seats, 3 rows
with 36 seats, 4 rows with 27 seats, 6 rows with 18 seats, 9 rows with 12 seats, 108 rows with
1 seat, 54 rows with 2 seats, 36 rows with 3 seats, 27 rows with 4 seats, 18 rows with 6 seats,
and 12 rows with 9 seats.
The arrangement I would use for displaying a message would be 6 rows of 18 seats because
messages need to be long width ways so that you can fit whole words in the arrangements. I
chose 18 x 6 instead of 12 x 9 because 12 by 9 would be more square-like than 18 by 6, which
won't be able to fit a whole phrase or message.
Reflection: I found it was easiest to find the factors of 108 to find the possible seating
arrangements because I have a strategy to do it. The strategy was to take the numbers from 1
to 12 and see if they are factors of the product by testing them with the divisibility rules.
Whichever numbers were factors, I would divide the product by to find their counterpart in
the factor pair. If I felt that some factors were missing, I would test 13, 14, and so on.