Math: Multiplication

How to do big multiplication fast?

Here's a fast and easy multiplication method that I currently use. Vedic and traditional methods, as well as, some other multiplication tips, tricks and strategies are fine, but this multiplication method requires absolutely no guessing, it's not difficult mental gymnastics and the numbers are readily known to anyone adept at (1–9) × (1–9) tables and proficient at adding and using compliments to add.

I'm only interested in basic math functions and this suits my needs, often I just do the work in my head and write the results as I go.

Multiply by combining. I can look at something like (213×7) and know it's 1491, without having written anythng, because (213×7) is (7×2) = 14 and (7×1) = 7 and (7×3) = 21 and I picture the 14 7 21 and separate any available tens places from available one's places to their left 14 (7 2)1 then add 14(9)1 to get the total 1491 I'd write it, like this, if I didn't just do it in my head and write the product as I went, which I won't do herein. :

What about additional carrying? (687×9) is much the same: 54 72 63 is 5(4 7)(2 6)3 is 5(11)(8)3 is 5(11)83, but what of that double digit 11? Carry it by combining again. (5 1)183 is (6)183 = 6183

Do the same with each digit in multi-digit factors, but tack on zeroes for place value holders at the end of the appropriate results and, then, you will add the results, together.

For instance, (345×7889), you do (3×7889) but 3 is actually 300, so you add two zeroes onto the end of the result, then you do (4×7889), but as 4 is actually 40, so you tack one zero onto the end of that result and, finally, do (5×7889) and 5 is actually 5, so there's no additional place value zeroes that are needed, so, now, just add all of the results.

Note: Subtract by adding. For instance, (35–4): (4+6) = 10 and (10+20) = 30 and (30+5) = 35 so 6+(20+5) = (6+25) = 31

Moving on to some exponent examples:

27(3)^2 - 4(3)^3

27(5)^2 - 4(5)^3

27(6)^2 - 4(6)^3

Bonus multiplication problem: 12345×67890 the vedic criss cross method would do this in one line, but requires too much thinking for my tastes, especially, when the factors get much larger. This took me hardly any time or effort.

First, I wrote the formula, then all the multiples, which were already known multiples, so, I didn't have to calculate anything, then I did all of the carrying, then I lined up and wrote the results with their place values and then I added them. Finally, I wrote the answer.

Easy peasy. Took less than a few minutes to do.

Could have been faster, but I'm still a learning and take awhile. I still use calculators to check my work.

We've used exponents, so, here's an easy way to solve exponents using binary. You can learn to Easily Convert Decimal to Binary , but for now, I'll just use this guide.

Find the binary of the exponent. For instance, in 8^6 the binary of the exponent 6 is 110, so, drop the leftmost binary digit to get 10, now, for each 1 put an S8 (S means we square and 8 means we multiply by 8, we use 8 in this example, because that is the number being raised by the exponent) for each 0 put an S (again, S means we square, which is just multiplying a number by itself). So the binary number 10 becomes S8S

“S” square 8 = 64

“8” multiply 64 by 8 = 512

“S” square 512 = 262,144

8^6 = 262,144

Here's how I would sloppily do it:

Another example: 12^9

Binary of 9 is 1001

Drop leftmost binary digit 1001 becomes 001

Assign S12 (the number being raised by the exponent) to the 1's and S to the 0's to get SSS12

“S” square 12 = 144

“S” square 144 = 20,736

“S” square 20,736 = 42,981,696

“12″ multiply 42,981,696 by 12 = 5,159,780,352

12^9 = 5,159,780,352

Here's how I would messily do it:

So you see, x^2 would simply be x squared and x^3 would simply be x squared and the result multiplied by x as I have done in your examples above.