Counting Pills
Express rules in words
See more than one rule for a given pattern
Devise and use problem solving strategies to explore situations mathematically
This problem is the second of four problems relating to triangular numbers. The others are Triangular Numbers, Level 3, Triangular and Square Numbers, Level 5, and Triangular Number Links, Level 6. Over the range of these problems we develop the idea of the triangular numbers leading to an algebraic formula for the nth triangular number.
This problem is also related to Building Patterns Incrementally, Algebra, Level 3.
This problem is about patterns, how to continue them and how to find the general term of a pattern. Many of the ideas that are used with this problem can be used with other patterns. So finding the next term (by looking at the recurrence rule – see Building Patterns Constantly, Algebra, Level 3), using a table, and incorporating number properties are all valuable skills that can be used in many situations.
Problem
Pharmacists sometimes use a triangular tray to quickly count pills. Make a tray out of cardboard that can be used to count up to 15 table tennis balls.

Teaching sequence
- Introduce the problem to the class. Brainstorm ideas for approaching the problem and keeping track of what has been done.
- Encourage them either to use a table and to see how the entries in the table are related or to use a geometrical approach. (The latter can be found below or referred to in Triangular Numbers, Level 3. It involves a little rearranging of the pills.)
- As the students work on the problem in pairs you might ask the following questions to extend their thinking:
What strategies might help you to find the answer?
How can you use your knowledge about numbers here?
Can you see any patterns that might help?
Can you put those patterns into words?
We give some more scaffolding questions in the solution. - Share the students’ answers. Ask them to explain their reasoning. Take special care that the words that they use for the rules are accurate. Encourage them to think about both approaches that we have used below.
Which approach can you understand best? - Ask students to write up a solution.
- Initially use the extension problem for the more able students but try to let the others have some time working on it. (You might want to use smaller numbers to make it easier. Try 66 to start them off. They can probably do this by drawing the triangular numbers. Encourage them, though, to use the word rule for triangular numbers that they have found.)
- Discuss this with the whole class.
Extension to the problem
How many pills will there be along the side of the triangular stack if there are 6216 pills altogether?
Solution
This problem has been solved by a range of simpler methods at earlier levels (see Triangular Numbers, Level 3). In this solution a slightly more sophisticated approach is taken.
Try multiplying across. Challenge the students to complete the table. What do they have to multiply by each time in order to go from the term to the triangular number?
(The entries in the third column should be, in order, 1, 1.5, 2, 2.5, 3, 3.5, 4.)
What is the pattern for the numbers in the third column? How do they relate to the other two?
How can you now get the third triangular number? The 10th? Can you give a rule for the triangular numbers for any given term? (Take the term and multiply by half of the next term.)
|
term |
triangular number |
multiply by? |
|
1 |
1 |
|
|
2 |
3 |
|
|
3 |
6 |
|
|
4 |
10 |
|
|
5 |
15 |
|
|
6 |
21 |
|
|
7 |
28 |
|
|
8 |
|
|
Help them to see that the number in the third column is half of the next term (4 is half of 8). So the 7th triangular number is 7 x (8/2) = 28. If this is applied to the 99th row of the table we will have to multiply by a half of 100. This leads to a solution of 99 x 50 = 4950.
|
term |
triangular number |
multiply by? |
|
1 |
1 |
|
|
2 |
3 |
|
|
3 |
6 |
|
|
4 |
10 |
|
|
|
|
|
|
99 |
99 x (100/2) |
100/2 |
|
100 |
|
|
|
|
|
|
This is actually more fun from a geometrical point of view. Put two lots of the 99th triangular numbers together. Colour one red and the other blue (see below). How big are the sides of the rectangle that is produced? (The rectangle is 99 by 100.) So 2 x (99th triangular number) = 99 x 100. This means that
the 99th triangular number = (99 x 100)/2 = 4950.

Students should be encouraged to express in words the relationships involved. For example, to say,
“a triangular number is the term number x (the next number term)/2”; or
“a triangular number is (the term number x (the term number plus 1))/2”.
Can you see that both of these numbers are the same?
Can you write the expressions using algebra? (This is not expected for Level 4 students but some of them might be able to understand it. You will need to use your judgement here.)
Extension:
We have 6216 pills. Now this is a triangular numbers and the term of that number is the length of the side of the pills. So
6216 = (the term number x (the term number plus 1))/2.
This means that
12432 = the term number x the term number plus 1.
Can you find two factors of 12432 that differ by 1 and that multiply together to give 12432? By experimenting or using square roots you should be able to find that 111 x 112 = 12432. So 6216 is the 111th triangular number.
| Attachment | Size |
|---|---|
| Counting Pills.pdf | 47.14 KB |
| CountingPillsMaori.pdf | 49.23 KB |
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