Galactic Matyhematics.

@George,

Quote:

Obviously, I can complete it.
Of course, I won't.
It would be tedious.
Your point is to show a trick to make the calculation quickly, is it not?
Go right ahead.
You have my full attention.

ok, well, it IS tedious if it is done the '' ol'fashioned way'. I agree.
It is awkard, tedious, cumbersome, errorprone and what have you!

Here is how it is done in VM (Vedic Mathematics):

We use the Sutra:

Quote:

By one more then the one before"

(Ekadhikina Purvena)

Here we go:

1/19=

we take the ''1' for the '9' and add '1', so we get 1+1=2, we gone use this '2'
here in our calculations.

we write down '1' :

1/19= 1

Now we multiply the '2' (the number we calculated above) with this '1'
and put it for the '1'. Like this:

1/19= 21

Then we calculate 2*2 (the two in the above row)=4 :

1/19= 421

And we go on like this and multiply the 4 with our '2' again. 2*4=8

1/19= 8421

Then we mutiply that 8 by our '2' again. 2*8=16 en we put the '6' in and remember the '1'

1/19= 68421

Then we multiply the '6' with our '2' , 6*2=12, but because we had to 'remember' the '1', it becomes 12+1=13. Now we put down the '3' in the row and remember the '1'.

1/19= 368421

Then we multiply the '3' with our '2' . 3*2=6 and we add the '1' we remembered.
so itr becomes '7'

1/19= 7368421

Then we mutiply the '7' with our '2;. 7*2=14 and we put down the '4' and remember the '1'.

1/19= 47368421

Then we multiply 4 with our '2' again. 4*2=8. we add the '1' we had to remember and we have 9

1/19= 947368421

Now we have 9 decimals and we need (19-1)=18

It becomes even easier now!

Now we going to write it like this

947368421

-------------
999999999

Now we have to fill in the gaps and then we get:

947368421
052631578
------------
999999999

Now we also have 052631578

and we put that before our first row, so it becomes:
0,052631578947368421

voila!

writing it down takes way longer than actually doing it!

Or do you still prefer the ol''fashioned way? Wink

1 Reply

Why does using that process for 1/17 give the exact same result?

1 Reply

@maxdancona,

Quote:

Why does using that process for 1/17 give the exact same result?

What do you mean??

1 Reply

@Quehoniaomath,

It is pretty clear

Replace 1/19 with 1/17 in your post (where you go into details on how you calculate 1/19). Then go through the exact same process step by step.

You will get the same result for 1/17 that you got for 1/19.

1 Reply

Now who can do 13211 /9 by VM?

1 Reply

@Quehoniaomath,

Quehoniaomath wrote:

Now who can do 13211 /9 by VM?

Only you Quehoniaomath... only you.

(that's why it is so funny)

1 Reply

@maxdancona,

Quote:

Only you Quehoniaomath... only you.

Well, no, you can too, because I have showed it here in the thread how easy you can do that too.
Or don't you read the postings? Wink

0 Replies

@maxdancona,

Quote:

Replace 1/19 with 1/17 in your post (where you go into details on how you calculate 1/19). Then go through the exact same process step by step.

You will get the same result for 1/17 that you got for 1/19.

I thought you ment the same outcome, hence my question.

1 Reply

@Quehoniaomath,

You get the same result, the same outcome, the exact same number as an answer.

That is why it is funny.

1 Reply

@maxdancona,

Quote:

You get the same result, the same outcome, the exact same number as an answer.

That is why it is funny.

??

I don't know what you mean now.

1/19=0,052631578947368421

1/17=0,058823529411764705

There IS a difference

AND I haven't written that you can calculate 1/17 the same way.

1 Reply

The given method for 1/19 can be used for 1/29, 1/39 1/49 with the right outcome!

1 Reply

Now, anyone here able to calculate x in:

Quote:

1/(x+4) + 1/(x+3)=1/(x+2) + 1/(x+5)

the conventional way? If you do please give the steps!

1 Reply

@Quehoniaomath,

It's so much easier if you use a slide rule.

0 Replies

@Quehoniaomath,

It clearly can't be used for 1/17.

2 Replies

@Quehoniaomath,

Your way (or Mr. Tirhaji's) is just as tedious and error-prone.
This is 2014.
Who does such calculations without a calculator,
aside from those doing parlor trick?

0 Replies

@Quehoniaomath,

No--but this I'd like to see

Start with

1/(x+4) + 1/(x+3)=1/(x+2) + 1/(x+5)

and manipulate algebraically

Combine numerators
((x+3)+(x+4))/((x+3)(x+4))=((x+2)+(x+5))/((x+2)(x+5))

Add numerators
(2x+7)/((x+3)(x+4))=(2x+7)/((x+2)(x+5))

divide both sides by 2x+7
1/((x+3)(x+4))=1/((x+2)(x+5))

multiply out to form quadratics
1/(x^2+7x+12)=1/(x^2+7x+10)

Invert
x^2+7x+12=x^2+7x+10

Cancel common terms
12=10

??????

Rap

2 Replies

@raprap,

Ahem. You divided by zero (2x+7).

0 Replies

@maxdancona,

For 1/17, start with 7 and use 12 as the multiplier. Method 3 here (http://en.wikibooks.org/wiki/Vedic_Mathematics/Sutras/Ekadhikena_Purvena) shows how to arrive at 7 and 12.

0 Replies

@maxdancona,

Quote:

It clearly can't be used for 1/17.

well, I didn't say you can!

0 Replies

@raprap,

Quote:

No--but this I'd like to see

Start with

1/(x+4) + 1/(x+3)=1/(x+2) + 1/(x+5)

and manipulate algebraically

Combine numerators
((x+3)+(x+4))/((x+3)(x+4))=((x+2)+(x+5))/((x+2)(x+5))

Add numerators
(2x+7)/((x+3)(x+4))=(2x+7)/((x+2)(x+5))

divide both sides by 2x+7
1/((x+3)(x+4))=1/((x+2)(x+5))

multiply out to form quadratics
1/(x^2+7x+12)=1/(x^2+7x+10)

Invert
x^2+7x+12=x^2+7x+10

Cancel common terms
12=10

??????

ok, thanks, It is wrong but I do appreciate the effort!
You just showed how clumsy and errorprone it is in the conventional mathematics.
Thank you very much

Here is the right conventional way:

Quote:

(x+4)(x+5)(x+2) / (x+4)(x+5)(x+2)(x+3) + (x+3)(x+5)(x+2) / (x+4)(x+5)(x+2)(x+3) = (x+3)(x+4)(x+2)/ (x+4)(x+5)(x+2)(x+3) + (x+4)(x+3)(x+5) / (x+4)(x+5)(x+2)(x+3)

Now:

(x+4)(x+5)(x+2) + (x+3)(x+5)(x+2)=(x+3)(x+4)(x+2) + (x+4)(x+3)(x+5)

We are going to calculate:

1. (x+4)(x+5)(x+2)

2 (x+3)(x+5)(x+2)

3 (x+3)(x+4)(x+2)

4 (x+4)(x+3)(x+5)

We start with:

1. (x+4)(x+5)(x+2)=(x^2+ 9x + 20) (x+2)=x^3 + 2x^2 + 9x^2+18x + 20x + 40=x^3 + 11x^2 + 38x +40

2 (x+3)(x+5)(x+2)=(x^2 + 8x + 15)(x +2 )=x^3 + 2x^2 + 8x^2 + 16x + 15x + 30=x^3 + 10x^2 + 31x + 30

3 (x+3)(x+4)(x+2)=(x^2 + 7x +12)(x+2)=x^3 + 2x^2 + 7x^2 + 14x +12x + 24=x^3 + 9x^2 + 26x + 24

4 (x+4)(x+3)(x+5)=(x^2 + 7x +12)(x+5)=x^3 + 5x^2 + 7x^2 + 35x + 12x + 60=x^3 + 12x^2 + 47x + 60

This means:

x^3 + 11x^2 + 38x +40 + x^3 + 10x^2 + 31x + 30 = x^3 + 9x^2 + 26x + 24 + x^3 + 12x^2 + 47x + 60

We add a lot now:

2x^3 + 21x^2 +69x + 70 = 2x^3 + 21x^2 + 73x + 84

that becomes:

69x + 70 = 73x + 84

that is the same as:

73x-69x=70-84

So

4x= -14

And

Quote:
x=-14/4=-3.5

Agreed ?

wasn't that cumbersome, long, boring and errorprone?

The good news is you can do this one in vedic mathematics (VM) within a few seconds!

1 Reply

Galactic Matyhematics.

Related Topics

Quick Links

My Account

able2know