I’d like to step up to 3x3 multiplication using this method he uses for 2x2:

65x34

65x30 + 4x65

1950+260

2.210

Is it possible?

I’d like to step up to 3x3 multiplication using this method he uses for 2x2:

65x34

65x30 + 4x65

1950+260

2.210

Is it possible?

I assume you mean Arthur Benjamin?

The answer is yes. However; the method you quote is not Arthur Benjamin’s.

Let me explain.

If you want to use Arthur Benjamin’s method, you want to do it this way:

65x34 =

34x65 = (lowest number first)

(34-4) x (65+4) + 4 x (65-30) = (first goes down, second goes up, add the difference: 4, times the other difference(65-30=35))

30 x 69 + 4 x 35 = (30 x 69: easier is: 30 x 70- 30)

(2100 - 30) + (140)

2100 + 110 =

2210

Now with 3x3 it becomes a bit more complicated, involved and maybe taxing on the brain.

An easy example:

343x657 =

340 x 660 + 3 x (657-340) =

340 x 660 + 3 x 317 = (I’ll show you how I calculate 34 X 66)

500^2 - 160^2 + 951 =

250,000 - 25,600 + 951 = (-25,600 + 951 = -25,000 - 600 + 1,000 - 49 = -25,000 +351)

250,000 - 25,000 + 351 =

225,351

For 34 X 66 I use difference of squares. You can also fall back to any other 2x2 method.

In the previous example, when we take the 343 down to 340, the 657 became 660. This effectively makes it a 2x2.

A more general example would go like this:

343x653 =

340 x 656 + 3 x (653 - 340)=

340 x (660 - 4) + 3 x 313 =

340 x 660 - 4 x 340 + 3 x 313 =

500^2 - 160^2 - (3 X (340-313) - 340) =

250,000 - 25,600 - 3 x 27 - 340 =

224,400 - 421 =

223,979

Is this what you mean?

Or are you looking for different ways of calculating 3x3’s?

How can I start mental calculating

I’ve got both Arthur Benjamin’s book and Bill Handley’s book. The multiplication technique quoted is not in the Benjamin book but it is throughout the Handley book where he uses his circle method to make it a little easier for non-Maths people to use.

Next month, I’m going to participate in the “Mental Calculation World Cup 2018” in Germany, where there is a specific announced Challenge task in the form of:

(3 digit x 3 digit ) + (8 digit / 3 digit)

e.g. (256 x 441) + (56052768 / 628)

To be honest for such an example of 3 by 3 multiplication, I’m gonna use my own Nodas’ method, since it’s just 1 line in my mind.

For Kinma’s example, I’m mentally calculating the line below (horizonally) :

343 x 653 = 18 | 9x43 + 30 | 2304-25 = 18 | 417 | 2279 = 223979

same with

256 x 441 = 8 | 82+224 | 2296 = 112896

(alternatively 256 x 441 = 336 squared = 112896, much easier)

If this doesn’t work well, like in hard examples such as 937 x 858, then the **horizontal criss-cross **should work whatsoever, no matter what. I currently average around 11 seconds (for each one) in my training for such 3x3 (screenshot from my training this week) , and I think I could go down to 7 or 8 seconds for each 3x3, if I train more.

Then I plan to commit this 5 or 6 digit result, to my short-term memory via my Major system PAO and simultaneously figure out the (8 digit over 3 digit) division.

Such (8 over 3) integer mental divisions like "56052768 / 628 " seem much easier than the ones I solved in Memoriad 2016 (where “10 digit over 5 digit” is the standard category.) and I got 6 out of 10 correct, in 6 minutes. So, timewise, such (8 over 3) it should be around 20 second for me. My mental plan for the above 56052768 / 628 is just five simple (3 over 2) steps :

560/63 = 8r56

(565+2x8) /63= 9r14

(142+2x9 ) / 63 = 2r34

(347+2x2) / 63 = 5r36

(366+2x5) / 63 = 6r(-2),

from the above, I get the correct 89256. If any doubt, the 6th step removes the doubt:

(-20+8+2x6 ) / 63 =** 0r0** .

Then add these 5 digits (89256) to the (5 or 6) digit result of the first part (3x3) to get the final (5 or 6) digit result.

89256 + 112896 (recall from my PAO) = 202152 (for which I needed another 3 seconds to figure out)

It’s kinda hard to work this out without any application of soroban or abacus, like all the Asians will do. I gotta be less mechanical and improvise more and that hopefully work out some correct results in that Challenge task which I mentioned.

Nodas