Wednesday, 9 May 2012

Fun with Numbers: some styles of manipulation

There are several ways of manipulations while dealing with number-problems. Trying out these ways is fun and helps in building math-analytical skills.
Consider a product of consecutive odd integers,
X = 3*5*7*9....47.
We observe a regular pattern of numbers spread over with a common difference between every two consecutive numbers. Yes, this is from a topic called 'Series and Progressions' of 'Algebra' and this is an arithmetic progression.
The number of terms = {(Final term-Initial term)/(common difference)}+1 = (47-3)/2 + 1 = 23

One thought goes like this:
Pair up the first and last terms, second and last-but-one, etc..Then all the terms get paired up except the middle one '25'.


If you observe it, sum of the two terms in each pair is constant and is 50.
3+47 = 50
5+45 = 50
23+27 = 50.

This is natural as the two terms in any pair are centred around the middle term 25 and hence have an arithmetic mean of 25.
Arithmetic Mean?, yes, now gears are changed to 'statistics'
(3+47)/2 = 25

"The two terms in any pair are centred around 25", this observation leads us to a little manipulation of each term in relation to the number '25'.

Again it's the turn of 'Algebra'. The idea that comes to the mind is
(a-b)(a+b) = a2-b2, apply it..


All the steps in a nut shell:
X = 3*5*7*9....47

The other thought goes like this:
Pair up every two consecutive terms starting from the first one. As there are odd number of terms, the last term remains unpaired.


The two numbers in every pair can be manipulated in terms of the mean of that pair. For example, take the pair (3,5). Its mean = (3+5)/2 = 8/2 = 4. The number 3 can be manipulated as '4-1' and 5 can be manipulated as '4+1'.
All the steps in a nut shell:
X = 3*5*7*9....47
    =  (3*5)(7*9)(11*13)...(43*45)(47)
    = {(4-1)(4+1)}{(8-1)(8+1)}.....{(44-1)(44+1)}(47)
    = (42-1)(82-1)(122-1)....(442-1)(47)

Let us see one problem here:
Find the sum : 2+4+6+...48
There is no need of remembering any formulae or typical math terms to solve this problem.
First let us find the number of terms. To find it, we can take out ‘2 common from all the terms,
This shows that there are 24 terms.
As we discussed in the first approach, build pairs of numbers with first and last, second and last-but-one,...
Each pair contributes to an equal sum of '50'. Now we can easily find the sum of all terms. Isn't it?

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