Memoriad sqrt

I’ve been training ten-digit square roots very regularly for over a year. After a break of 2 months, I started again and immediately set a personal best. If anyone is still training in this category?

With best regards, ynnad


Amazing! :slight_smile:

Might I ask about your multiplications and additions results?

И моят син се занимава активно и има чудесни ризултатш

[Translated with Google Translate: And my son is active and has great results]




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Hello ynnad wich method do you use for that?

Hello, which method do you use to get that achievement which is under the world record.

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I haven’t trained the six-figure for a long time. For that I need something under 100 seconds. But under 60 seconds is really extremely good. What method does your son use?

I only practice Flash Anzan on a regular basis. Multiplication up to six digits x five digits and division in a similar range.

I use the following method.


Не е много добър на умножение и събиране.Използва кръстосано умножение.Не използва соробон.Даниел Тиимс описва подробно метода в този форум

From time to time, I train the square roots of four-digit numbers with a precision of 8 digits.
My average time is between 20 and 25 seconds.
I use a similar method to ynnad but with a two digit divisor.


Използва базов метод,нищо специално.

Of course, it also has advantages that you immediately know the first two digits. I’ve also used this variant for a while.

Your son seems to have really great abilities.

What exactly is meant by the Basic Method? Anything can be meant by that.

But now a topic that has bothered me for a long time. Why do some mental calculators keep their methods so secret? Are they afraid that others could also use their “unknown” method and then maybe even be better in the end? To date, nobody knows exactly which method was used by the Indian kids for square roots. Why does it all have to be so opaque?


Main advantage is that I have a much lower chance of getting a negative remainder from the division.
Knowing the squares up to 100 allows to get 2 digits immediately.
Additionally, if we optimize the order of calculations, we can use the knowledge of squares a second time!

Basic method with a two-digit divisor
(1-2) 5178-71^2=137==>137x5=685 (in the first step it is x5 instead of x10)
(3) 685+(3:2)=686,5==>686.5:71=9 r:47,5
(4) 475+(6:2)-9^2/2=437,5==>437.5:71=6 r:11,5
(5) 115-9x6=61==>61:71=0 r:61
(6) 610-9x0-6^2/2=592==>592:71=8 r:24
(7) 240-9x8-6x0=168==>168:71=2 r:26
(8) 260-9x2-6x8-0^2/2=194==>194:71=3(in the last step, the result of the division is closer to 3 than to 2)

Optimized algorithm ( works only for an accuracy of 8 digits)

The first 5 digits are calculated in the same way
(1-2) 5178-71^2=137==>137x5=685 (in the first step it is x5 instead of x10)
(3) 685+(3:2)=686,5==>686.5:71=9 r:47,5
(4) 475+(6:2)-9^2/2=437,5==>437.5:71=6 r:11,5
(5) 115-9x6=61==>61:71=0 r:61
In step 6, half of the square (6^2/2) was omitted
(6) 610-9x0=610==>610:71=8 r:42
Later, the middle product (6x0) was omitted.
The last 3 digits of the result are currently 608.
Divide by 10, round off and subtract half a square.
420-9x8=348==>3480-61^2/2=1600 (here, an approximate result is enough)
(7-8) 1600/71=23


I find your optimized algorithm very interesting.


(1-2) 51 - 49 = 2 ==> 2 x 5 = 10

(3) 13.5 ÷ 7 = 1 r: 6.5

(4) 69 - 0.5 = 68.5 ÷ 7 = 9 r: 5.5

(5) 56.5 - 9 = 47.5 ÷ 7 = 6 r: 5.5

(6) 58 - 40.5 - 6 = 11.5 ÷ 7 = 0 r: 11.5

(7) 115 - 54 - 0 = 61 ÷ 7 = 8 r: 5

(8) 50 - 18 - 0 - 8 = 24 ÷ 7 = 2 r: 10

(9) 100 - 0 - 72 - 2 = 26 ÷ 7 ≈ 3

Actually, a two should come out. Through a lot of experience I have developed a feeling for how big the remainder could be in the next step. Therefore I assume that the 6th decimal place is ≥5.

Of course, I’m not always right about that. Can you apply your estimate for the last two digits to my algorithm? Actually, I only need one step more than you, but it still doesn’t seem that easy to transfer.

If I simply leave out the inner multiplication in step (7), I never get to the 8. With your variant, of course, that doesn’t matter because the remainder are larger due to the two-digit division.

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Well, if you skip the inner term in step 8 instead of 7 you get
50 - 0 - 8 = 42
In my variant, this is the remainder of the division in step 6
The next steps are

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Метода е това, което показва Камил или подобно.Със сигурност знае квадратите от 1÷99 и повече.
Индийските деца ползват същия метод плюс ведическа математика.Имат изключително силно обучение.Даниел Тиймс има подробно обяснени техники ,същите по които работи моето дете.Той е треньор в тази област.

Thank you very much, I tried it out a bit right away. It works great but unfortunately no alternative for me. It’s probably just a matter of automation. Nevertheless, I am certainly a bit slower in the end. It’s like anything, you can’t have one without the other.

By the way, I basically use the same algorithm for cube roots (similar to how Daniel Timms describes it on his homepage). At the beginning you can divide the difference between the first 3 radical digits and the next cube number by 3. Thus the divisior and the by-products are one place smaller.