How to do this in 10 seconds without a calculator and scratch paper?
1565 x 1197 = ?
Call it 1565 * 1200 which is just (1565 * 2 *100) + 1565*1000
The subtract 1565 three times.
Do you know how to apply Indian's classic calculation method to this ?
Sorry never heard of it.
Ask google - it takes less than 5 seconds!
I don't think that'll work.
I googled 1565 x 1197, and got the answer: 1 873 305.
It didn't take 10 seconds. Then I googled "India's classic calculation method" but I got nothing I could use.
Jekertee, are you willing to teach us the method?
Here's a question: how randomly chosen is that pair of numbers? After you demonstrate the method, I want to see you do the same thing with numbers provided by an audience member.
:shy , Jim, I din't know that method also, I made a quizz thinking that someone knew to tell me too
I input random numbers into the search box of google but found only strange result. It's amazing to me the example I gave works on your search
Only numbers? Didn't you specify an operator?
What about an expression like (into the search box):
2*17.9 + 3**3
Is there any information you could give us to help search for it?
1565 x 1200 - (1565 x 3) = ?
4 x 1565 = 6000 + 240 + 20 = 6260
6260 x 300 - (1565 x 3) = ?
6260 x 3 = 18000 + 600 + 180 = 18780
18780 x 100 = 1878000 - (1565 x 3)
1565 x 3 = 4500 + 180 + 15 = 4695
1878000 - 4695 = ?
This is one way, and it's probably not the fastest, but it uses an ancient technique of multiplying where one number was halved, and the other was doubled. Or multiplied by 3, and divided by 3.
it's called vedic math. if you google it, you'll find a ton of sites with math tricks here and there. but usually, they don't involve numbers that big.
when multiplying 2, 2-digit numbers with the same tens digits and two ones digits that add up to 10, (ex. 23 x 27 ) the vedic stuff is useful.
you just multiply the tens digit by the next number up, so multiply 2 by 3 and get 6.
then multiply the two ones digits, and get 21.
the answer is 621.
there's tons of little things like that that you can find on google.
Yes, here's a proof. Let a be the common tens place digit (2 in this case), and b one of the ones place digits (either 3, or 7 in this case). Then
[tex](10a + b)(10a + 10 - b)[/tex]
[tex]= 100a^2 + 100a - 10ab + 10ba + 10b - b^2[/tex]
[tex]= 100a(a + 1) + b(10 - b)[/tex]
That is to say, multiply a by the next number up, multiply by 100 and add the product of the two numbers in the respective ones places.
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