HCF with Bottle Caps

Introduction: Teaching aids other than chalk and blackboard attract children and extend their attention span so that they are able to grasp basic concepts, especially in Maths. Waste materials that are easily available are very useful in making ideas clearer to children. They also make learning interesting and joyful. For example, one can use bottle-tops of different colours to introduce the concept of factors and to teach HCF, and even LCM. Let's see how we can do this.

Objective: 

• To help children learn by doing

• To make children appreciate the value of maths

• To build confidence in children that maths can be fun to learn

Activity Steps: 

HCF with bottle tops

Teachers of maths often find it difficult to create an interest in the subject in tender minds. This, however, is vital to teaching and to help children understand mathematical concepts. The activity method is perhaps the most effective. If one starts a lesson with an activity, the students not only enjoy the lesson but also understand the concept better.

Session 1

Step 1: Let’s make rectangles

Give each of the children in your class a specific number of bottle tops.Ask them to arrange them in all possible rectangular arrays.The bottle tops should be arranged in rows. Single straight lines, both horizontal and vertical, are also acceptable.Each time a rectangle is formed, the number of rows is a factor of the given number.When it is not possible to arrange the given bottle tops in a rectangular form, shift bottle tops one by one from the end to form a new row.

For example, to find all possible factors of six we follow the steps below:

                     Fig. 1.a shows six bottle tops in a single row. So 1 is a factor of 6.

 Fig.1b shows two rows with 3 tops in each row. A rectangle is formed, so 2 is a factor of 6. The figure that follows has 3 rows with 2 tops in each row. Since a rectangle is formed with three rows, 3 is a factor of 6.


In Fig. 1.d. a rectangle is not formed, so this should be ignored.Figure, 1.e has two bottle tops in first row and one in each of the remaining rows. No rectangle is formed, so 5 is not a factor of 6. The next possible rectangular arrangement is a vertical line with six rows. So 6 is a factor of 6.

Prime or not?
We can also ask the children to check whether a given number is prime or not using this method. The reason is that no rectangular form other than a single horizontal or vertical line is possible for a prime number.

Step 2: The highest common factor (HCF)
Now let us see how to find the HCF of two given integers.Use two sets of bottle tops, each of a different colour, for the two given numbers.Arrange the two sets of bottle tops on either side of a demarcation line: for convenience, we may take the smaller number of bottle tops on the left and the larger on the right.Arrange the bottle tops on the left in a vertical line.Ask the children to justify this arrangement. The reason, as we know, is that HCF is always lesser than or equal to the greatest factor of the smaller number.Now arrange the ones on the right in as many rows as there are on the left. If you are able to arrange these in a rectangle, then the number of rows in the arrangement is the required HCF.If such an arrangement is not possible, change the arrangement on the left to the next possible rectangular arrangement, and then accordingly change the arrangement on the right as well.When both the sides are in a rectangular arrangement with the same number of rows, the number of rows is the HCF.

For example, let us try to find the HCF of two numbers, 6 and 20.In the first step, we arrange 6 bottle tops in a vertical line to the left of the demarcation line (Line of Control!). We now arrange the 20 other bottle tops on the right. We find that we cannot arrange these in a rectangle (Fig. 2.a).So 6 is not the HCF of 6 and 20.


We now change the arrangement on the left to three rows of two bottle tops each.Then we change the arrangement of the 20 bottle tops.Again we find that these do not form a rectangle (Fig. 2.b).The next step is to change the arrangement on the left yet again into two rows of three bottle tops each.Now we find that we can arrange the 20 bottle tops in two rows.Therefore 2 is the HCF of 6 and 20 (Fig. 2.c).In case of three numbers, we use two demarcation lines and three differently coloured sets of bottle tops.

This idea can be extended to teach LCM too, if we spare a little time to think.