Manual square and cube opening

I. Preface

When I was in junior high school, the math teacher taught us to manually open the square. At that time, I was shocked. I verified it several times and found that the method was really good! Now I am a graduate student. I recall that the methods for manual square construction have been vague. I checked Baidu today and found that this method is from Chapter 9 arithmetic. This method was already found in China thousands of years ago!

2. Manual square opening

Chapter 9 is described as follows:

Square example:
Set product to real. Take a calculation. Step by step. Super 1. Income. Take a multiplication by a number as the formula. In addition. Except already. The doubling method is fixed. Its Division. Fold. The next step is to reset the lending process. By review. Secondary income. In addition. In addition. Starting with the income pair. Complex division fold as before.
If it is not enough, it cannot be opened. If there is actually a score, the entire score is set. It is Kai Zhi, Bu, Kai his mother report division. If the mother is not allowed to open, and then set by the mother, but it is open, the end, such as the mother and one.

2.1 Algorithm Description

1. divide the integer part of the square number from the beginning to the left into two segments, separated by an asterisk, into several segments, indicating that the square root is the number of digits; the fractional part is separated from the two digits after the highest bit. The accuracy + 1 is required for the number of segments.

2. Calculate the number on the highest bit of the square root Based On the number in the first segment on the left.

3. Subtract the square of the highest bit from the number of the first segment, and write the second segment on the right of their difference to form the first remainder.

4. multiply the number of the highest bits obtained in the second step by 20 to try and remove the first remainder. The obtained maximum integer is used as the test operator.

5. Multiply the 20 times of the maximum number of digits obtained in step 2 by the operator. If the product is less than or equal to the remainder, the tester is the second digit of the square root. If the obtained product is greater than the remainder, the tester is reduced and tried again, the first operator smaller than the remainder is the second number of square root. (That is, 3 is the second digit of the square root.

6. In the same way, continue to calculate the number of others on the square root. Use the remainder minus the product obtained in the above method to form a new remainder with the array in the third segment. In this case, ask the tester and multiply the first two digits of the square root obtained above by 20 to try the new remainder. The maximum integer obtained is the new tester ..

7. the tests for new testers are as follows.

2.2 example:

1. Split each two digits from the single digit to the left into segments.

2. The highest bit is 10, and the number of digits is not greater than 10, that is, 3

3. 10-3 ^ 2 = 1, remainder 1

4. 3*20 = 60, trial provider 134/60 = 2

5,134/(60 + 2) = 2, more than 10, less than 62, so it is determined that the second digit is 2

6. 32*20 = 640, trial provider 1026/640 = 1

7, 1026/(640 + 1) = 1, more than 385, so it is determined that the third digit is 1

8. Calculate the next bit at a time. The remaining 0 is known.

3. Manual opening:

Chapter 9 arithmetic description:

Chapter 9 arithmetic:
Question 9: There are 1,860,867 feet today. Is it cube?
Answer: One hundred and twenty-three.
Opening method:
Set product to real. Take one calculation step, which is second-class. The income of the discussion is calculated by multiplying the borrowed one, and the result is excluded. Except, the three statements are fixed. Distinct, fold down. Set the number of rows in a three-digit number. Let's take another example. In the next step, the second class is in the middle of the second class. Reset, take a multiplication, and then take a multiplication. Division by definite method. Except "already", "Times", and "and" center. Complex division, fold down as before.
It cannot be opened. If the score is accumulated, the sum of the points is set. In fact, it is open, and its mother is opened to report the division. If the mother cannot start, and then multiply by the mother, it is open. , Such as the mother and one.

3.1 Algorithm Description

1. Divide the integer part of the number of opened cubes into a group from the beginning to the left;

2. Based on the leftmost group, obtain the maximum number of digits of the cube root;

3. Subtract the cube with the highest digits of the cube root from the first group and write the second group on the right;

4. Divide the remainder by 300 times of the square of the obtained highest digit to obtain the trial operator;

5. write the 300 times of the obtained highest-digit Square and the product of the trial operator, 30 times of the obtained highest-digit square, and the square area of the trial operator and the cube of the trial operator on the left side of the vertical layout, check whether the sum is greater than the remainder. If the sum is greater than, reduce the operator and try again. If the sum is not greater than the sum, the operator is the second digit of the root;

6. In the same way, continue to calculate the number of other members of the root. Tests for new testers are also performed in the preceding method.

3.2 Example