A previous article briefly introduced the multiplication method of the Trachtenberg system, and the address is here. Trachtenberg also developed faster calculations for certain numbers.
Let’s start with the quick method of multiplying by 11. It is calculated according to what we might call the neighbor rule:
From right to left, add each digit and the adjacent digit to its right and take its units place. If the value is greater than 9, the tens place is carried to the next place (the carry is at most 1).
So in the future, when you multiply by 11, you just write it, for example:
For example, 633 x 11:
1: there is no number on the right side, so I call it 3. The rule is derived here, that the first digit of the value is equal to the first digit of the multiplicand.
Number 2:3 + 3 = 6
Number 3:6 + 3 = 9
So, 0 + 6 = 6
When calculating, it is customary to add a 0 to the multiplied number, which is more pleasing to the eye:
In the example above, the sum of two adjacent numbers has never been greater than 9, so I decided to do another chestnut and calculate 1754 x 11. When the sum of two adjacent numbers is greater than 9, we can easily mark the result with a small dot, such as:
Number 3:7 + 5 = 12, so the number 4:1 + 7 + 1 = 9.
How’s that？ Simple enough?
The multiplication rule for 12 is as simple as for 11:
Multiply each digit by 2 and add the value of its neighbor to the ones place. If the value is greater than 9, the tens place is carried to the next place.
Let’s go straight to chestnuts: 413 x 12
1. (from right, same below) : 3 x 2 + 0 = 6. For the convenience of calculation, we add a zero in front of the multiplier, so that for beginners, the calculation of the last digit will not be easily ignored. Veteran words, direct imagination can be, after the operation of the front fill 0, no longer do special instructions.
1 x 2 + 3 = 5
4 x 2 + 1 = 9
0 x 2 + 4 = 4
How’s that？ Or write the result directly when you see the number, which is much easier than traditional calculation, right?
As usual, we have to solve the carry problem caused by multiplying by 2 plus neighbors. Since the maximum value of this operation is only 27 (9 x 2 + 9), the maximum carry value is 2.
One more chestnut: 63247 x 12
1:7 x 2 + 0 = 14, leave 4, carry 1. Remember in the last video we put a little dot in front of it to indicate that we have a carry?
4 x 2 + 7 = 15 + 1 = 16 Leave the 6 and carry the 1.
2 x 2 + 4 = 8 + 1 = 9
3 x 2 + 2 = 8
6 x 2 + 3 = 15 Leave the 5 and carry the 1.
0 x 2 + 6 = 6 + 1 = 7 Note that if the value is greater than 9, the carry is written directly to the next 1 bit.
Now we’re going to multiply the numbers 5, 6 and 7.
If you have the book, please turn to page 28.
Let’s start with 6. Damn, why do I have to start with six? Please look down.
There are two rules for multiplying 6. Let’s start with the first one:
Add each digit of the multiplicand to the right half of the adjacent digit, keeping the ones digit, and carry the tens digit to the next digit if the value is greater than 9.
The operation of taking half of the “right adjacent half” is exactly called “taking half and rounding”, which is to take half of the integral part of the odd number 1, 3, 5, 7 and 9. For example, if half of 5 is 2.5, we just take 2, and so on.
Go ahead: 622084 x 6
Position 1:4 + 0/2 = 4
The 2nd bit: 8 + 4/2 = 10, take 0, carry 1.
Position 3:0 + 1 + 8/2 = 5
Position 4:2 + 0/2 = 2
Number 5:2 + 2/2 = 3
Sixth place: 6 + 2/2 = 7
Position 7:0 + 6/2 = 3
However, that’s not the whole rule for multiplying by six, the whole rule is:
Add each digit to the right half of the adjacent digit. If the number is odd, add 5 first. The resulting value is reserved in the ones place. If the value is greater than 9, the tens place is carried to the next place.
Go straight to chestnuts: calculate 443052 x 6
1: I ran out of numbers on the right, so I got 2
Second place: notice that this is a 5, add 5 first, then add half of the right side, 5 + 5 + 2/2 = 11, keep the 1, carry the 1 in the tens place
Position 3:0 + 1 + 5/2 = 3. Note the half-and-round operation of 5/2. In addition, it is recommended to get into the habit of adding digits first when there is a round, such as saying “1” mentally when you see a 0 in this example.
Number 4:3 is odd, so we have to add 5:3 + 5 + 0/2 = 8
4 + 3/2 = 5. Note that a good habit to develop in practice is not to think “half of 3 is 1, 4 plus 1 is 5”. Ideally, think “4,5” in your mind, or “4,1,5” in the early stages of practice.
Number 6:4 + 4/2 = 6
Position 7:0 + 4/2 = 2
If you look at this, you might say, well, there’s no advantage over the traditional algorithm, right? And it’s more complicated, right?
It looks like this! In the traditional algorithm, the multiplication formula is used to multiply two numbers, and then process the carry.
But IN my opinion, the biggest advantage of Trachtenberg’s algorithm is the simplicity of carry, because the maximum is only one. It’s also relatively easy to convert multiplication into addition.
At this point, there are melon people pointed out that you this shit drop and division, dare to say simple?
Pro, take half of the operation hot yao simple, do you also afraid?
Now please turn to page 35, and we will begin to learn the multiplication rule for seven.
The rule for multiplying 7 is very similar to that for 6:
Multiply it by 2 and add half of its right-hand neighbor. If the number is odd, add 5 after multiplying it by 2. The resulting value is reserved in the ones place. If the value is greater than 9, the tens place is carried to the next place.
Let’s look at chestnuts: 3412 x 7
Position 1: No data on the right, so multiply by 2, get 4
Number 2:1 is odd, so multiply by 2 and add 5. 1 x 2 + 5 + 2/1 = 8
4 x 2 + 1/2 = 8
3 x 2 + 5 + 4/2 = 13.
Number 5:0 x 2 + 1 + 3/2 = 2. Make a habit of adding bits first
And finally, let’s do the multiplication rule for 5. It’s a bit like a cross between rules 6 and 7, but simpler:
Take half of the right adjacent of each digit of the multiplicand add 5 if the current digit is odd
We don’t even have to worry about rounding, because if we take the largest number 9 in half, we get 4, and if we add 5, we get 9.
As usual, we use a chestnut to understand this rule: calculate 436 x 5
The first digit: 6 is even, and there are no numbers to the right of it, so it’s written as 0.
Number 2:3 is odd, 5 + 6/2 = 8
Third digit: 4 is even, so just take 3/1, get 1
Bit 4:0 is even, so just take 4/2, get 2
So just to summarize, in the multiplication of 5, each digit of the multiplicator is just used to determine whether or not to add 5, it’s not involved in the operation. So it’s a lot easier to calculate.