The Accident

I have about a month or less until I go back to work, which means my maternity leave is coming to an end soon.  No, let me make a correction.  It’s my EXTENDED maternity leave.  I was really hoping that Justin Trudeau’s extended 18-month maternity leave would kick in before I returned.  But I think I may have to wait to have another kid in order to take advantage of that – which is most likely not going to happen.  Don’t worry, after the last birth, I don’t think I’m too thrilled to do it again.  Besides, I am turning 43 next month.

To keep my math teaching skills sharp, I tutor once a week.  Grade 8 Math.  The last session we were working on the concepts of Factors and Multiples.  Students who have a difficult time with this usually confuse the two with each other.  We worked on a really great problem which I took from the 7 ONAP.  ONAP stands for Ontario Numeracy Assessment Package. It is essentially a diagnostic tool that is used to assess students’ prior knowledge, based on the previous grade.  It allows teachers to plan and differentiate teaching strategies to help students get to the grade level.

Here’s the problem.

The Accident.  There are 24-100 glass bottles on the shelf.  When placed in rows of 6, the remainder is 3.  When placed in rows of 8, the remainder is 7.  How many glasses were on the shelf?  Is there more than one possible answer?

This problem has multiple entry points and allows for many possible solutions.  I can also see how students would not know where to begin – immediately feeling confused after reading the information.  This is also a very good example of how literacy is involved because you do have to understand what the information is telling you.

In any case, when we are told that there are remainders when the glasses are placed in rows of 6 and 8, immediately, we should be thinking that the total number of glasses cannot be a multiple of 6 or 8.  This means we can eliminate all the numbers that are multiples of 6 and 8 right away.

I had my student list out all the multiples of 6 and 8.  Then list all the numbers from 24 to 100.  From there, we eliminated all the multiples of 6 and 8 and looked at what was left. Here is what it may have looked like. The highlighted numbers are the multiples of 6 and 8.


From here, we would test out each number and see if they are the possible number of glasses by adding 3 to each multiple of 6, and adding 7 to each multiple of 8. This method may be considered a lower entry level way of solving the problem.

Another student may decide to list all the multiples of 6 and 8 that are in between 24 and 100 and add the remainder of 3 and 7, respectively.  Then proceed to compare and look for the same numbers in each list.  Like so.


As you can see, this method is much more efficient as it deals with the multiples and the remainders right away.  The highlighted numbers are the possible numbers of glasses that are on the shelf.  And there are three possible numbers – 39, 63, and 87.

Upon further observation, one will notice that the difference between these three numbers is always 24.  Why 24?  Well, the least common multiple of 6 and 8 is 24 and so it would make sense that the next number would have to be a multiple of both 6 and 8.

Here, I have presented two solutions.  Can you think of another way to solve this, that may involve a different strategy?  Perhaps, a higher level?  Leave me a comment and let me know what you think.


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