Jo's Table

Problem

Jo has some squared paper handy. She put the even numbers from 2 to 12 along the top and the numbers 3, 6, 9, 12, 15, 18 down the side. She then started adding the numbers together. As she filled in the numbers she began to see patterns.

Find some patterns for yourself. Use them to complete Jo’s table.

Teaching sequence

1. Introduce the problem to the class. Ask them
   What number do you think goes here?
   And what about here?
   Put these numbers onto the table.
2. Let the students work on the problem with a partner. Check that they understand what they are supposed to be doing.
3. As a whole class share the students’ solutions. Make sure that they are looking for patterns along the diagonals. Ask them to say why their pattern works.
4. Encourage them to go on to try the Extension problem.
5. Discuss the answers that the students get in a class situation. Ask them why their answers work. Possibly ask them to make up questions of their own.

Extension to the problem

What is the smallest number that appears more than once?
 

What is the largest number that appears only once?
 

How many times does 15 occur?
 

If the table is continued for ever across and down, how many times will 40 appear? Which rows and columns is it in? (Note: This extension will challenge students working at levels 3 & 4.)

Solution

Jo’s completed table is shown below.

There are a large number of patterns that can be found here. The horizontal patterns are even and odd numbers. These occur because Jo is adding even numbers together or even numbers plus an odd number. Diagonal patterns such as the one in red, go up in 5s. This is because the horizontal numbers are increasing in 2s and the vertical numbers in 3s. Back diagonals (such as the one in blue) only increase in 1s. This is because there is an increase of 3 downwards but a decrease of 2 to the left. 3 – 2 = 1. Encourage the class especially to look for patterns that occur in the diagonals.

Solution to the extension

It’s worth noting from the start that every number outside the red border appears at least twice. This can be seen by looking at the columns and noting the repetitions between them. A quick scan shows that every number inside the red box appears only once. So the smallest number that appears more than once must be the smallest number outside the box. This is 11.

From what we have just said, the largest number that appears only once must be the largest number inside the orange box. This is 12.

The number 15 only appears in two columns, so it occurs twice.

How would we get 40? First of all 40 could not occur to the right of the ‘40’ column nor could it occur below the ‘39’ row. So what numbers in Jo’s table will up to 40? We’ll go systematically across the columns.

2 + ? = 40, ? = 38 but is not a multiple of 3;
4 + ? = 40, ? = 36 and 36 is a multiple of 3; √
6 + ? = 40, ? = 34 but is not a multiple of 3;
8 + ? = 40, ? = 32 but is not a multiple of 3;
10 + ? = 40, ? = 30 and 30 is a multiple of 3; √
12 + ? = 40, ? = 28 but is not a multiple of 3;
14 + ? = 40, ? = 26 but is not a multiple of 3;
16 + ? = 40, ? = 24 and 24 is a multiple of 3; √
18 + ? = 40, ? = 22 but is not a multiple of 3;
20 + ? = 40, ? = 20 but is not a multiple of 3;
22 + ? = 40, ? = 18 and 18 is a multiple of 3; √
24 + ? = 40, ? = 16 but is not a multiple of 3;
26 + ? = 40, ? = 14 but is not a multiple of 3;
28 + ? = 40, ? = 12 and 12 is a multiple of 3; √
30 + ? = 40, ? = 10 but is not a multiple of 3;
32 + ? = 40, ? = 8 but is not a multiple of 3;
34 + ? = 40, ? = 6 and 6 is a multiple of 3; √
36 + ? = 40, ? = 4 but is not a multiple of 3.

So 40 appears six times in Jo’s table.

This can be found more efficiently in two ways. First it is quicker to go through the systematic approach above but using 3 + ? = 40, 6 + ? = 40 and so on.

But if we note that 40 = 20 x 2 and that 2 x 3 = 3 x 2, we can write down the following ways of getting 40. Neglecting the first one of these we see another way that we can get 40 in six ways.

40 = 20 x 2;
40 = 17 x 2 + 2 x 3;
40 = 14 x 2 + 4 x 3;
40 = 11 x 2 + 6 x 3;
40 = 8 x 2 + 8 x 3;
40 = 5 x 2 + 10 x 3;
40 = 2 x 2 + 12 x 3.