# Investigate calendars, and look for any patterns.

Extracts from this document...

Introduction

Maths Coursework

Introduction: I was given a task to investigate calendars, and look for any patterns. I noticed several patterns, the first of which was the relationship between the starting days of different months, also I noticed the relationship between numbers in columns of the calendar, the relationship between numbers in the rows, Studying diagonal relationships, and Studying relationships between adjacent numbers.

1. Days on which months start

First, I explore the days on which different months start.

Which months are the same/ have a pattern?

Ex 1.1 Study Sample Calendar:

Month | Starting day |

1 | Friday |

2 | Monday |

3 | Monday |

4 | Thursday |

5 | Saturday |

6 | Tuesday |

7 | Thursday |

8 | Sunday |

9 | Wednesday |

10 | Friday |

11 | Monday |

12 | Wednesday |

From the above table, I can see that some of the months start on the same day, which means there may be a pattern when compared with other years. If so, then that means there is a pattern of which months start on the same day each year. The results of this first test are as follows:

1, 10 = same

2, 3, 11 = same

4, 7 = same

5

6

8

9, 12 = same

Now I must investigate to find out if the pattern is the same in other years. In order to do this, I check a calendar of the year 2004.

## Ex 1.2: Results for 2004

Months ( n ) | Start day |

1 | Thursday |

2 | Sunday |

3 | Monday |

4 | Thursday |

5 | Saturday |

6 | Tuesday |

7 | Thursday |

8 | Sunday |

9 | Wednesday |

10 | Friday |

11 | Monday |

12 | Wednesday |

Therefore, we can see that the dates are thus:

1, 4, 7 = same

2, 8 = same

3, 11 = same

5

6

9, 12 = same

10

Middle

3. Exploring the relationship between numbers in the rows

Now I explore the horizontal numbers…

Ex 3.1 The first row in June:

n | |

1 | 1 |

2 | 2 |

3 | 3 |

4 | 4 |

5 | 5 |

The formula is obviously n = n here, but will it change in the next few rows?

Ex 3.2 The second row in June:

n | |

1 | 6 |

3 | 7 |

4 | 8 |

5 | 9 |

6 | 11 |

7 | 12 |

As you can see from the table above, the formula is quite simple:

n = n + 5

Ex 3.3 The third row in June:

n | |

1 | 13 |

2 | 14 |

3 | 15 |

4 | 16 |

5 | 17 |

6 | 18 |

7 | 19 |

The above table gives another formula: n = n + 12

Ex 3.4, Finally the fourth row in June:

n | |

1 | 20 |

2 | 21 |

3 | 22 |

4 | 23 |

5 | 24 |

6 | 25 |

7 | 26 |

The final formula is n = n + 19

Ex 3.5:These are the formulae for the rows in the month of June:

First row: n = n

Second row: n = n+5

Third row: n = n+12

## Fourth row: n = n+19 I didn’t draw the table for 5th, but predict it as:

Fifth row: n = n+26

The general expression is n = n + x depending on the start date.

4. Studying diagonal relationships

Ex 4.1 Diagonally descending to the left, starting with 2nd January.

n | date |

1 | 2 |

2 | 8 |

3 | 14 |

4 | 20 |

5 | 26 |

The difference is always 6. n x 6 gives the sequence:

6 12 18 24 30, but to get the correct sequence subtract four:

n = 6n – 4

I will test the formula on 2 other diagonals descending to the left: one more January, one July.

Ex 4.2: Diagonally descending to the left, starting with 9th January.

n | date |

1 | 9 |

2 | 15 |

3 | 21 |

4 | 27 |

## The above table gives the formula n = 6n + 3

Ex 4.3: Diagonally descending to the left, starting with 5th June.

n | date |

1 | 5 |

2 | 11 |

3 | 17 |

4 | 23 |

5 | 29 |

Conclusion

1 | 2 | 3 | 1 x 24 = 24 3 x 22 = 66 Difference = 42 |

8 | 9 | 10 | |

15 | 16 | 17 | |

22 | 23 | 24 |

Finally, I decided to test a box of 14 numbers.

Ex.5.15 Starting 1st August

1 2 3 4 5 6 7 | 1 x 14 = 14 |

8 9 10 11 12 13 14 | 7 x 8 = 56 |

Difference = 42 |

Ex.5.16 Testing a box of 16 numbers, starting 1st June.

1 | 2 | 3 | 4 | 1 x 25 = 25 |

8 | 9 | 10 | 11 | 4 x 22 = 88 |

15 | 16 | 17 | 18 | Difference = 63 |

22 | 23 | 24 | 25 |

Ex.5.17 A box of 18 numbers.

1 | 2 | 3 | 4 | 5 | 6 | 1 x 20 = 20 |

8 | 9 | 10 | 11 | 12 | 13 | 6 x 15 = 90 |

15 | 16 | 17 | 18 | 19 | 20 | Difference = 70 |

Ex.5.18 A box of 20 numbers

1 | 2 | 3 | 4 | 5 | 1 x 26 = 26 |

8 | 9 | 10 | 11 | 12 | 5 x 22 = 110 |

15 | 16 | 17 | 18 | 19 | Difference = 84 |

22 | 23 | 24 | 25 | 26 |

I decided to plot a table summarising the above results:

## Ex. 5.19 | |

## Number of figures in box | Difference between multiplied diagonals |

4 | 7 |

6 | 14 |

8 | 21 |

9 | 28 |

10 | 28 |

12 | 42 |

14 | 42 |

16 | 63 |

18 | 70 |

20 | 84 |

Note that the above table includes the 7 x table, which is once again significant as there are seven days a week. Therefore, there is a relationship between the figures involved in calendar calculations.

However, I must comment on this last table, Ex 5.19– it’s odd that certain numbers you would logically expect to be in the sequence are not present, as you never get the numbers 35, 49, 56 or 77. All these numbers are multiples of 7, and yet they do not follow an orderly sequence as you increase after 9 figures in a box: the pattern is not regular. I am unable to work out why this is so.

Areas of possible further work

I could have explored the following:

- Division of diagonals, but this would be difficult because they would not be whole numbers.

- Addition or subtraction of diagonals in boxes, e.g.

1 2 1 + 9 = 10

8 9 2 + 8 = 10 - Addition of columns, adding rows, e.g.

1 + 8 + 15 + 22 - A write up of whether the last day of each month follows the same pattern as the first month, in each year except leap years, i.e. an investigation similar to that of Ex 1.

This student written piece of work is one of many that can be found in our GCSE Consecutive Numbers section.

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month