6 cellos, 12 violas, and 21 violins Jo wants them to sit in rows each section the same number of rows.

Question

6 cellos, 12 violas, and 21 violins Mr. Jo wants them to sit in rows, and each section to have the same number of rows. What is the greatest number of rows the orchestra would have?

Answer 1

This is just a greatest common factor problem, with a real-world application which is nice for a change. Most word problems require operations and most GCF problems just have you name the GCF among three numbers. That can leave you wondering when you would ever need to know GCF in the real world.

Also, I love that these proportions are actually realistic for a typical orchestra class.

Anyway, this is a very simple problem that a lot of people can do pretty much instantly if they have had sufficient practice, but I am going to break it down for the fun of it.

Factors of 6: 1, 2, 3, 6

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 21: 1, 3, 7, 21

The greatest common factor is 3, so that answers the question.

Now, while the question is realistic, it would be a mistake to assume that because there are three rows, we can divide each section by three to determine how many people would be in each row. That is beyond the scope of the question and unlike the question does not line up with real-world conventions. I am going to continue Mr. Jo's problem because reality is that the answer to the GCF question is not going to be the end of his calculations.

The standard in most orchestras is to have curved rows, not straight rows. The front row generally has eight people: four violins, two violas, and two cellos.

21 - 4 = 17 violins remaining

12 - 2 = 10 violas remaining

6 - 2 = 4 cellos remaining

The cellos work themselves out nicely so that there are two in each row, and that's perfect because musicians share music stands with a partner, so we want them in pairs except maybe the very last stand. That means we cannot have five violas in the remaining two rows because that would require an extra stand. We need even numbers in both of those rows, instead. We have no choice about putting the 21st violin player on their own stand, but we can otherwise divide the remainder of the violin section somewhat evenly.

Second row:

17 - 8 = 9 violins remaining

10 - 4 = 6 violas remaining

We can put all of the remaining instruments in the third row.

So, we have the same number of rows for each section like Mr. Jo wants, and because the rows are curved, we have fewer people in the front than in the middle and fewer people in the middle compared to the back. We also didn't unnecessarily use an extra music stand by making sure everyone except the last violin has a stand partner.