Modern Approach to Speed Math Secret & Trachtenberg

(Andrew Penton) #1

I have read a bit about the book “Modern Approach to Speed Math Secret” By Vitthal Jadhav and it seems like a very good book on mental calculation. I have previewed a bit of it and the method it provides for squaring four-digit numbers is really quite nice. However, it seems like it might be a bit advanced for me. Can anyone who has it give me a recommendation on whether or not it would be suited for someone who is essentially a beginner to mental calculation?

I am also interested in how Pascal’s triangle can be used when calculating powers/roots. I have only known it to be useful for powers of 11, but it seems like it can be useful for other numbers like 1.01. Can someone explain this concept or point me to a book that does? Is it discussed in the Modern Approach?

#2

Vitthal used to be on this forum. Years ago.

I indeed own and have read this book. It is great for learning about mental calculation!
However; the learning curve is steep.
Vitthal explains things that are sometimes useful only in very narrow cases. So if you want to learn about mental calculation in general and are a beginner, start with a different book. This one for example:

Pascals Triangle is indeed in his book.
He uses it for expanding (a+b)^n, where n is a row in the triangle.

If we take for example a third power, then the third row is [1 3 3 1].
In general a number (a+b)^3 expands to:
1a^3 + 3a^2b + 3ab^2 + 1b^3

This can be done on any number. In your case of 1.01, a=1 and b = 0.01.

Hope that helped!

#3

BTW, I discussed this book 5 years ago with a user called ‘Rock’, which I always thought was the writer himself

Might be interesting to read. See here:

(Andrew Penton) #4

Thank you very much, this is exactly the feedback I wanted.

#5

There is Jakow Trachtenberg, who published “High Speed Maths”. This is a good start to beginn with.

(Andrew Penton) #6

I’ve only been able to find “The Trachtenberg Speed System of Basic Mathematics”. Is this what you’re referring to? If not, could you link it? I’ve heard a lot about the Trachtenberg system and it seems very interesting.

EDIT: After looking more at the Trachtenberg System, it seems more like a written system than a method for mental calculation. Is there a way to adapt it to better suit mental calculation? I find that it’s very simple, just not applicable for solving problems in your head.

#7

You must drill yourself to become an Maths-Magician. All mental calculations are based on written workouts and careful observations of what happens. You must put some energy into the learning process.
The book from Anne Cutler is really “The Trachtenberg Speed System of Basic Mathematics”.
Here you have 2 Link for further studies:

https://www.researchgate.net/publication/236871340_Rapid_Mental_Somputation_System_as_a_Tool_for_Algorithmic_Thinking_of_Elementary_School_Students_Development

You will meet Harry Lorayne, Arthur Benjamin, Donald Ervin Knuth, …

Watch the references !

#8

Please describe exactly what you want to achieve.
Some people just want to be able to calculate without machines, but with paper and pen present.
Others, myself included, like the challenge of using only the brain, so the techniques can be used in traffic, while waiting on a train, etc.

The latter needs techniques for keeping the amount of numbers in short term memory small for example.

I have read this book decades ago. Back then it did not click with me. Might be a personal thing or I may have missed something back then.

#9

Ok, I did.

The Trachtenberg system of speed mathematics.

These are my personal observations, based on my personal way of calculating. I encourage everybody interested in speed calculation to read the book, doing the exercises and then see if the method works for you.

What I like about the book:

• checking yourself through 9-proof and 11-proof. If you read my posts here on this forum you know that I love and advocate to check myself using the same methods. I use a slightly different way to do the 11-proof, which I describe in my posts.

What I don’t like about the book:

• too many rules to memorize. Only at the end of the book is it explained why the system works the way it works. I would rather see this in the beginning of the book.
• this might be a personal thing, but I would forget all the rules if I don’t practice them often. My brain works like this; if I can recreate the rules based on the knowing the system, then it is much easier to recreate the rules after a long period of non usage.
• I advocate learning to calculate from left to right and this book mostly works from right to left, unless it is division or square roots, which can not be done right to left. The rules can be tweaked to multiply from left to right btw. If anybody is interested we can do this on this forum.

(Andrew Penton) #10

I was hoping to just do calculations completely mentally, but thank you for noting this. It makes a lot more sense why people would use Trachtenberg now.

This is exactly the problem I’ve had with this method, so if you could explain it that would be great. Although it seems like just starting at the other end of the number and accounting for any carries should do the trick… Is this what you’re referring to or is there something I’m missing that would be easier?

#11

Ok, so let’s modify Trachtenberg for people that calculate from left to right.

For example multiplication with 8. This one has the most rules.
These are the rules for right to left:

1. First figure: subtract from ten and double.
2. Middle figures: subtract from nine and double what you get, then add the neighbor.
3. Left-hand figure: subtract two from the left-hand figure of the long number.

Let’s see how and why this works and for simplicity’s sake, let’s take a 2 digit number:
If we have a 2 digit number with the first digit ‘a’ and the second ‘b’, we can write this as:
a | b
where the | separates the digits.
If we multiply a|b with 8, using the rules above we get:

a-2 | 2(9-a)+b | 2(10-b)

If ‘a-2’ now denotes the hundreds, etc. we can also write this as:
100(a-2) + 10(2(9-a)+b) + 2(10-b) =
100a - 200 + 180 -20a +10b + 20 -2b =
80a +8b = 8(10a+b) =
8(a|b)
So yes, the rules indeed will multiply by 8.

Let’s check with an actual 2 digit number:
78 * 8 =160;
7-2 | 2*2+8 | 2(10-8) =
5 | 12 | 4 = (take care of the carry)
624

11-proofing:
First the starting point. 78 - 77 = 1. 1 * 8 = 8.
Now the answer: 624 - 550 = 74. 74 - 66 = 8. Alternatively: 4-2+6 = 8.
Both end in 8, so the answer is probably correct.

Some caution.
1: The Trachtenberg rules are great for digits which are close to 10.
With digits close to zero the rules still work, however imho they don’t speed up the calculation anymore. This is true whether you calculate from left to right or the other way round. An extreme example is multiplying 11 by 8:
11 * 8 =
-1 | 2 * 8 + 1 | 2 * 9 =
-1 | 17 | 18 = (now the 2 carries)
0 | 8 | 8 =
88

2: the system is designed for calculating from right to left as it keeps the amount of digits in short term memory as small as possible.
It is designed to put digit by digit on paper.

789 * 8 =
Following the steps we get for the units digit: 2(10-9) = 2.
tens: 2(9-8)+9 = 11 = 1 (carry the 1)
hundreds: 1+2(9-7)+8 = 13 = 3 (carry the 1)
thousands: 1+7-2 = 6
Result: 6312

3: The system is designed for multiplication with numbers until 12.
Above 12 Trachtenberg switches to what he calls Direct Multiplication. This is the exact same as criss-cross multiplication.

In those cases, add Arthur Benjamins system:
24 * 36 is a lot of work doing criss-cross.
However; this is where Benjamin shines:
24 * 36 = 20 * 40 + 4 * 16 = 800 + 64 =
864.

11 proof:
24-22 = 2 . 36 - 33 = 3. 2*3 = 6
864 - 770 = 94. 94 - 88 = 6. Alternatively: 4-6+8 = 6.
Both end in 6; answer is right.

My personal thoughts. The book states that the system is great for people with calculation problems. Being able to just work out a difficult calculation digit by digit can indeed give a student the confidence it needs to overcome this.

So this is a great thing.

#12

Let’s do a typical Trachtenberg multiplication by 8. However, left to right.

8 * 77,889

First a ballpark answer. 77,889 = almost 80,000. 8*80,000 = 640,000.
We overshot so the answer needs to be slightly lower than 640,000.

Using the Trachtenberg rules we get:
8 * 77,889
1: 7-2 = 5
2: 2(9-7) + 7 = 11. Add to 50 to get 61
3: 2(9-7)+8 = 12. add to 610 to get 622
4:2(9-8)+8 = 10. add to 6,220 to get 6,230
5: 2(9-8)+9 = 11. add to 62,300 to get 62,311
6: 2(10-9) = 2. add to 623,110 to get 623,112.

11 proof:
77889 mod 11 = 9 ( I hope you immediately see that 77,880 mod 11 = 0).
8*9 = 72. 72 mod 11 = 6. (just subtract 66 from 72).
So the answer mod 11 also needs to be 6.

So let’s do 623,112 mod 11.
I do this mentally, stripping numbers from the left, making the number smaller and smaller:
623,112 - 550,000 = 73,112 (mentally I just focus on the first 2 digits: 62. I just subtract 55 from 62)
73,112 - 66,000 = 7,112 (again; only focus on the first 2 digits: 73)
7,112 - 6,600 = 512
512 - 440 = 72
72 - 66 = 6