First, when I watched this thingy I was like UGHHH tratchenberg is soooo harddd but then I focused. I became the fastest multiplier in class omg thank u so muchh
Well, you can use a mnemonic system. I learned to do this from another source, Bill Handley's Speed Mathematics. Personally I visualize three places in my head, and one holds the answer, another holds the carry digits, and the last holds temporary results. I could tAlk more about the method if you asked me. Of course, you can go to the source material by going to Libgen.is and searching "Speed Mathematics".
But what about carry digits? For example, if you were to do 96* 55 left to right you might say "9*5 tens, 4". Then you'd say "9*5 ones, 5, 9*other 5 tens, 4, 6*5 tens, 3, add to get 12", so the correct thing to say would be 52, but you already said 4 and wouldn't be able to add the carry 1 to 4 to get the 5 in 52.
My grandfather could easily multiply two five-digit numbers in his head, and multiply much larger numbers by three or four digits. When I was in elementary school (In the late 90's), I asked him how he did it, and it he gave me the book The Trachtenberg Speed System of Basic Mathematics. I found the book fascinating and still use many of the tools in the book, especially the "rules" for multiplying large numbers by single digits. I also really liked his method for adding many large numbers together quickly with built-in error checking. It's really cool to see so many people interested in this because of Gifted, I'll have to check out the movie. I highly recommend anyone interested to look up the book.
In Vedic Mathematics, this is called "Vertically and Crosswise." However you call it, this is amazingly efficient. Those claims about multiplying two huge numbers together in a matter of seconds? Proven. Solid method.
I was soo confused but it's actually easy. All you do is just multiply, take the higher place digit, and add it to the next multiplication set. Easy Work!
Didn't know this is now used by a lot of people and became very famous as I look through the comments. I was thought this when I was 8 and I had a slightly hard time learning the whole system. But to tell you, once you etched this system onto your head, math would be a lot lighter. So, don't give up and research a bit more!
Oh my gooodddd I’m thirteen and they should’ve really taught us this in math! This is sooo easy I thought it was going to be hard, I think the only really hard part is trying to do this in your head and not having to write it down.
As far as I understand, this is not the trachtenberg system. You don't have to worry about the whole product in each intermediate step. Just Unit place with current number and Ten's place with previous digit. Thus nick named UT method. Let's take an example solved in the video. 562 x 37 Before starting the multiplication prefix a ZERO infront of the multiplicand. So it will look like, 0562 x 37 will make use of this ZERO later. Multiplication start with '7', the unit place of the multiplier, 37. Now '7' reaches the unit place of multiplicand, that is '2'. We call '2' as the Current Number (CN) for '7'. Any number previous to current number in multiplicand is called the Previous Number (PN). At this point, '7' don't have any PN. When multiplying with CN, take only UNIT place. That is '2 x 7 => 14', but we take only unit place, ie. 4. Add Ten's place of the product with the PN. Since there is no previous number, nothing to be added. U(2 x 7) + T() => 4 . . 0562 x 37 ---------------- 4 '7' travels to next number, in the multiplicand, that is '6'. Now '6' becomes the Current Number (CN) for 7 and '2' becomes Previous Number (PN). Unit place with CN and Ten's place with PN. That is, U(6x7) + T(2x7) => 2 + 1 => 3 ----------------(A) Similarly, '3' the ten's place digit in multiplier, 37, starts it's journey with the unit place '2' (of multiplicand 562). Remember, for '3', the CN is '2' and there is no PN. Unit(2x3) + Ten() => 6 ----------------(B) Add both (A) and (B) => 3 + 6 => 9 . , , . 0562 x 37 ---------------- 94 Now '7' jump to next digit in multiplicand, that is '5'. For '7' CN is '5' and PN is '6'. Similarly '3' also jump to it's next number and for 3 CN is '6' and PN is '2'. U(5 x 7) + T(6 x 7) => 5 + 4 => 9 --------------(A) U(6 x 3) + T(2 x 3) => 8 + 0 => 8 ---------------(B) Add both (A) and (B) => 9 + 8 => 17 => (1)7 We have a carry forward '1' and write the carry in brackets. . , , . 0562 x 37 --------------- (1)794 Now '7' reaches ZERO the final UT calculation point for it. So for '7' CN is ZERO and PN is '5' Our '3' reaches '5' which is it's CN and '6' becomes PN U(ZERO x 7) + T(5 x 7) => 0 + 3 => 3 -----(A) U(5 x 3) + T(6 x 3) => 5 + 1 => 6 ------------(B) Adding (A) and (B) => 3 + 6 => 9 Add the carry forward from the previous step, that is '1' 9 + 1 => 10 => (1)0 . , , . 0562 x 37 --------------- (1)0794 Calculations with '7' is DONE. Now '3' reaches ZERO. For 3, CN is ZERO and PN is '5' U(ZERO x 3) + T(5 x 3) => 0 + 1 => 1 Add the carry, that is 1 + 1 => 2 '3' is also DONE. As there is no more digits to follow (in multiplier), our final answer will be, , , . 0562 x 37 ------------------ 20794
Salve, purtroppo non parlo inglese, ho seguito le Vostre lezioni con molto piacere, e ne ho dedotto un avanzamento culturale. Peccato della lingua. Con fatica, mi son fatto aiutare da Google, ed anche da Google, ho tratto beneficio. Ho deciso di apprendere la lingua Inglese, così posso seguire le Vostre lezioni. Tank you
When the number becomes more than 2 digits (right side) I seem to be always missing one number in the equation. I've followed this video correctly but how am I always missing the the last number of the equation but get everything else right? Can someone help me out?
If your trying to learn this your should learn the addition method first. It might seem stupid because it seems simpler to do it the normal way but it will help you better understand how to do the harder stiff
I don't mean to appear rude but isn't that simply basic multiplication everyone learns in elementary? I mean, how else are you supposed to multiplicate without a calculator? Yeah, I'm here because of Gifted, too but I didn't know this method was called Trachtenberg method. I was hoping for something...I don't know, something you don't necessarily learn in school. Anyways, still thank you for the effort of doing a video.
1st, trachtenberg method isn't learn in school. 2nd, this meethod is teach so you wouldn't need a calculator or you'll able to compute bigger numbers using your head. 3rd, trachetenberg method isn't necessesaily learn in school
@@elysianzenith8000 How come I've learned it in school, then? Just because you use a claim two times, doesn't make it stronger. Also, I know this method is used, so you wouldn't need a calculator. That wasn't even the point, so why do you mention it?
Niyo-pun u come here because of Gifted right? "Trachtenberg System. Let you compute large number using your head.... So, u didn’t need a calculator" -chris evans. That’s why I said it
@@elysianzenith8000 But it has absolutely nothing to do with my comment. I didn't question the purpose of Trachtenberg method, so it's unnecessary mentioning it.
I can explain it. 123X456 has to be changed to look like this 000123X456 The last number has to be multiplied by the last number which would be 3X6 then you just multiply 4 and 5 by zero because of the pattern.basically it would looks like this 00012300X456 Pattern would be 0X4=0 0X5=0 3X6=18 So your first number is 8 and the 1 will be added to the second number which is 0X4=0 3X5=15 2X6=12 then you add them up with the 1 you carried and it would be 28 and your numbers would be 88 and you carry the 2 etc
0X9=0 6X2=12 carry the 1 Stage2 6X9=54 8X2=16+1+54=71 carry the 7.current numbers are 12 Stage3 9X8=72 7X2=14+7+72=93 carry the 9. Current numbers are 312 Stage4 7X9=63 2X0=0+63+9=72 carry the 7 .current numbers are 2312 Stage 5 are just multiplying 0 0X0=0 0X0=0+0+7=7.answer is 72,312
I’m here because of the movie Gifted I loved math and love working on equations so I thought maybe I was gifted BUT NOT I guess UPDATE I started using this method
Yeah, I see it but I don’t see how it’s easier than what I was taught in elementary school 40 years ago. I’m quite sure it’s easier than the shit they’re teaching now - using an entire piece of paper to do one multiplication or addition problem.
