# Algebra - Date Patterns.

Maths Algebra Coursework – Date Patterns

## Maths Algebra Coursework – Date Patterns

The purpose of this investigation is to analyse the relationship between patterns of dates and how changing variables affect these patterns. In order to examine these relations I will be creating formulas with the data available. Because numeric trial and error can be utilised an infinite number of times but never prove a theory completely, it is necessary to express the formulas in algebraic terms. The modifications that I will apply to the original anomaly will be:

1. Varying the number of columns within the grid itself consequently changing the figures in the actual box
2. Changing the box size to a grid of 2 (horizontal) by 3 (vertical)
3. Changing the box size to a grid of 3 by 3

My first task is to recognise an algebraic term for the basic pattern. For instance:

August

I have chosen to extract this box from the month of August given in the example on the original task sheet. I will use the root number of four throughout the investigation as a primary example in order to maintain continuity.

The pattern is as follows:

4 x 12 = 48

5 x 11 = 55

55 – 48 = 7

Instantly I recognised a relationship but I tested this again with a different set of values:

15 x 23 = 345

16 x 22 = 352

352 – 345 = 7

Under a preliminary analysis the data, when processed through the sums, seemed to give an undeviating result. The value was seven. Why did the operation keep resulting in the number seven, no matter what data I processed through it? My earlier recognition of a potential relationship was now strengthened; it seemed to me that the pattern would produce a value that was equal to the number of columns involved in the grid as a whole. In order to prove this I would have to create an algebraic formula for this particular event.

First I assembled an algebraic representation of the box:

n = root number

Using this I could concoct a generic formula that could be used with every box in a calender to find the difference. First I found separate formulas for each individual sum:

n x (n + 8) or n2 + 7 x n + n

This gives me the result of the root times it’s diagonal, eg:

4 x 12 = 48

4 x (4 + 8) = 48

(n + 7) x (n + 1) or n2 + n + 7 x n + 7

This gives me the result of the root plus one and it’s diagonal, eg:

5 x 11 = 55

(4 + 7) x (4 + 1) = 55

Now I have the two functioning formulas that form the basis of the pattern I can combine them in order to work out a formula that can be used for any root number. Fusing these formulas resulted in:

(n + 7) x (n + 1) – n x (n + 8) = d

or

n2 + n + 7 x n + 7 – (n2 + 7 x n + n) = d

d = difference

eg:

4 x 12 = 48

5 x 11 = 55

55 – 48 = 7

(4 + 7) x (4 + 1) – 4 x (4 + 8) = 7

42 + 4 + 7 x 4 + 7 – (42 + 7 x 4 + 4) = 7

or

20 x 28 = 560

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