1 .10 times 10:

Formula: head joint, tail to tail, tail to tail.

For example: 12× 14=?

Solution: 1× 1= 1.

2+4=6

2×4=8

12× 14= 168

Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space.

2. The heads are the same and the tails are complementary (the sum of the tails is equal to 10):

Formula: After a head is added with 1, the head is multiplied by the head and the tail is multiplied by the tail.

For example: 23×27=?

Solution: 2+ 1=3

2×3=6

3×7=2 1

23×27=62 1

Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space.

3. The first multiplier is complementary and the other multiplier has the same number:

Formula: After a head is added with 1, the head is multiplied by the head and the tail is multiplied by the tail.

For example: 37×44=?

Solution: 3+ 1=4

4×4= 16

7×4=28

37×44= 1628

Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space.

4. Eleven times eleven:

Formula: head joint, head joint, tail to tail.

For example: 2 1×4 1=?

Answer: 2×4=8

2+4=6

1× 1= 1

2 1×4 1=86 1

5. 1 1 times any number:

Formula: head and tail do not move down, middle and pull down.

For example: 1 1×23 125=?

Answer: 2+3=5

3+ 1=4

1+2=3

2+5=7

2 and 5 are at the beginning and end respectively.

1 1×23 125=254375

Note: If you add up to ten, you will get one.

6. Multiply a dozen by any number:

Formula: The first digit of the second multiplier does not drop, the single digit of the first factor multiplies each digit after the second factor, and then drops.

For example: 13×467=?

Solution: 13 bit is 3.

3×4+6= 18

3×6+7=25

3×7=2 1

13×467=607 1

Note: If you add up to ten, you will get one.

7. Multiple digits multiplied by multiple digits

Formula: The former factor multiplies each bit of the latter factor one by one, the second factor multiplies 10 times, the third factor multiplies 100 times, and so on.

For example: 33* 132=?

33* 1=33

33*3=99

33*2=66

99* 10=990

33* 100=3300

66+990+3300=4356

33* 132=4356

Note: If you add up to ten, you will get one.

A fast algorithm of "the sum of the first ten and the last ten is the same" and "the sum of the last ten and the first ten is the same" about the multiplication of two digits in mathematics. The so-called "from the end to ten" means that two numbers are multiplied, ten digits are the same, and the sum of single digits is 10. For example, 67×63, all ten digits are 6, and the sum of single digits 7+3 is exactly equal to 10. I told him that the multiplication of numbers like this is actually regular. That is, the product of the single digits of two numbers is the last two digits of the number. If it is less than 10, the ten digits are supplemented by 0; Multiply one of the ten digits by 1, and the result is the thousand and hundred digits of the number. Specific to the above example 67×63, 7×3=2 1, which is the last two digits of the number; 6×(6+ 1)=6×7=42, which is the first two digits of the number. Taken together, it is 67×63=422 1. Similarly, 15× 15=225, 89×8 1=7209, 64×66=4224, 92×98=90 16. After I told him this quick secret, the little guy was already a little excited. After pestering me to give him all the questions I could, and all the calculations were correct, he clamored for me to teach him the fast calculation method of "ending with the same top ten" I told him that the so-called "ten tail with the same head" means that two numbers are multiplied and the digits are exactly the same. The sum of ten digits is exactly 10, such as 45×65, and both digits are 5. The result of ten digits 4+6 is exactly equal to 10. Its calculation rule is: the product of two numbers with the same digits is the last two digits of the number, and if it is less than 10, the tenth digit is added with 0; Multiply dozens of digits and add the same single digit, and the result is hundreds and thousands. In the above example, 45×65 and 5×5=25 are the last two digits of the number, and 4×6+5=29 is the front of the number, so 45×65=2925. Similarly,11× 91=1001,83×23= 1909, 74× 34 = 2516,999.

In order to make it easy for everyone to understand the general law of two-digit multiplication, here are concrete examples to illustrate it. By comparing a large number of two-digit multiplication results, I divide the two-digit multiplication results into three parts, one digit, ten digits, more than ten digits, that is, hundreds and thousands. (The maximum multiplication of two digits will not exceed 10000, so it can only reach thousands. Now, for example, 42×56=2352.

Among them, the method to determine the single digits of a number is to take the mantissa of the product of two digits as the single digits of the number. Specific to the above example, 2×6= 12, where 2 is the mantissa of the result and 1 is a single digit;

The method to determine the decimal digits of a number is to multiply the individual digits of two numbers by the sum of the decimal digits respectively, and add the mantissa of the sum of the individual digits, which is the decimal digits of the number. Specific to the above example, 2×5+4×6+ 1=35, where 5 is the number of decimal places and 3 is the number of decimal places;

The rest of the number is the sum of the decimal places of two numbers and the product of decimal places, which is the hundred or thousand digits of the number. Specific to the above example, 4×5+3=23. Then 2 and 3 are one thousandth and one hundredth of a number, respectively.

Therefore, 42×56=2352. Another example is 82×97. According to the above calculation method, first determine the unit number, 2×7= 14, then the unit number should be 4; Then determine the number of decimal places of the molecule, 2×9+8×7+ 1=75, and the number of decimal places of the molecule is 5; Finally, the remainder of the number is calculated, 8×9+7=79, so 82×97=7954. Similarly, with this algorithm, it is easy to get the product of all two-digit multiplication.

Fast Calculation 4: Fast Calculation of Conditional Special Numbers

Fast calculation skills of two-digit multiplication

Principle: Let two digits be 10A+B and 10C+D respectively, and their product is s, which is expanded by polynomial:

S = (10a+b) × (10c+d) =10a×10c+b×10c+10a× d+b× d.

Note: Below, "-"stands for ten digits and one digit, because the number obtained by multiplying the ten digits of two digits is followed by two zeros. Please don't forget that the first product is the first two digits, the second product is the last two digits and the middle product is the middle two digits.

A. Fast multiplication

First, the top few are the same:

1. 1. The ten positions are 1, and the positions are complementary, that is, A = C = 1, B+D = 10, s = (10+b+d) ×/.

Method: One hundred digits are two, one digit is multiplied, and the number is the last product, and the first one is full.

For example: 13× 17

13+7 = 2-("-"is used as a mnemonic when you are not proficient, but you can use it when you are proficient)

3 × 7 = 2 1

-

22 1

That is, 13× 17= 22 1.

1.2. The decimal digit is 1, and the digits are not complementary, that is, A = C = 1, B+D ≠ 10, s = (1b+d) ×/kloc-.

Methods: The digits of the first multiplier are added to the second multiplier to get the first product, and the digits of the two numbers are multiplied to get the last product, which is full of ten and the first one.

For example: 15× 17

15+7 = 22-("-"is used as a mnemonic when you are not proficient, but you can use it when you are proficient)

5 × 7 = 35

-

255

That is 15× 17 = 255.

1.3. Ten bits are the same and the bits are complementary, that is, A = C, B+D = 10, and s = a× (a+1)×10+b× d.

Methods: Add 1 to ten digits, and multiply the sum by ten digits. The number is the front product, the number is multiplied by single digits, and the number is the back product.

For example: 56 × 54

(5 + 1) × 5 = 30- -

6 × 4 = 24

-

3024

1.4. Ten bits are the same, but the bits are not complementary, that is, A = C, B+D ≠ 10, s = a× (a+1)×10+a× b.

Methods: The first two multiplications, the number is the first product and the number is the last product. Multipliers add up, depending on their size, multiply the first of several multipliers by ten, and vice versa.

For example: 67 × 64

(6+ 1)×6=42

7×4=28

7+4= 1 1

1 1- 10= 1

4228+60=4288

-

4288

Method 2: Multiply the first two digits (that is, find the square of the first digit), the number obtained is the front product, the sum of the two mantissas is multiplied by the first digit, and the number obtained is the middle product. When the decimal number is full, multiply the two mantissas, and the number obtained is the back product.

For example: 67 × 64

6 ×6 = 36- -

(4 + 7)×6 = 66 -

4 × 7 = 28

-

4288

Second, after the same number:

2. 1. One bit is 1 and ten bits are complementary, that is, B = D = 1, a+c =10s =10a×/kloc-0+/kloc-0.

Methods: Multiply ten digits to get the product, and add 10 1.

- -8 × 2 = 16- -

10 1

-

170 1

2.2. < not very simple > the unit is 1, and the ten digits are not complementary, that is, B = D = 1, a+c ≠10s =10a×/kloc-0+.

Methods: The product of ten digits plus the sum of ten digits is the front product, and the unit is 1.

For example: 7 1 ×9 1

70 × 90 = 63 - -

70 + 90 = 16 -

1

-

646 1

2.3 bit is 5, and ten bits are complementary, that is, b = d = 5, a+c =10s =10a×10c+25.

Method: The product of ten digits plus the sum of ten digits is the front product, plus 25.

For example: 35 × 75

3 × 7+ 5 = 26- -

25

-

2625

2.4<'s not very simple > the unit is 5, and the ten digits are not complementary, that is, b = d = 5, a+c ≠10s =10a×10c+525.

Methods: Multiply two digits (that is, find the square of digits), the number obtained is the front product, the sum of twenty digits is multiplied by one digit, the number obtained is the middle product, and when the digits are full, multiply the two mantissas to obtain the number obtained is the back product.

For example: 75 ×95

7 × 9 = 63 - -

(7+ 9)× 5= 80 -

25

-

7 125

2.5. The positions are the same and the ten positions are complementary, that is, B = D, a+c =10s =10a×10c+b100+B2.

Methods: Ten digits multiplied by ten digits plus one digit is the front product, and one digit is added to the square.

For example: 86 × 26

8 × 2+6 = 22- -

36

-

2236

2.6. One is the same and ten are not complementary.

Methods: Multiply ten digits by one digit, the number is the front product, and add one square, and then see how much the sum of ten digits is greater or smaller than 10. Add a few bits to multiply a large number by ten, and vice versa.

For example: 73×43

7×4+3=3 1

nine

7+4= 1 1

3 109 +30=3 139

-

3 139

2.7. Non-complementary speed algorithm with the same number of digits and ten digits 2

Method: multiply the head by the square of the head and the tail, plus the result of multiplying the head and the tail by the tail and then multiply it by 10.

For example: 73×43

7×4=28

nine

2809+(7+4)×3× 10=2809+ 1 1×30=2809+330=3 139

-

3 139

Third, the special type:

3. 1, the number of a factor is the same from beginning to end, and the ten digits of a factor are multiplied by two digits with complement.

Method: Add 1 to the first place of complement.

For example: 66 × 37

(3 + 1)× 6 = 24- -

6 × 7 = 42

-

2442

3.2. The number of a factor is the same from beginning to end, and the ten digits of a factor are multiplied by two digits that are not complementary to each other.

Methods: Add 1 to the first digit of the random number, and multiply the sum by the first digit of the multiplicand. The number is the front product, and the two mantissas are multiplied, and the number is the back product. If there are no ten digits, add 0. Then see how much the sum of non-complementary factors is larger or smaller than 10, and multiply several numbers with the same number by ten, and vice versa.

For example: 38×44

(3+ 1)*4= 16

8*4=32

1632

3+8= 1 1

1 1- 10= 1

1632+40= 1672

-

1672

3.3. The numbers of a factor are complementary from beginning to end. Ten numbers of a factor are multiplied by two numbers with different digits.

Methods: Add 1 to the first digit of the multiplier, and then see how much the tail of different factors is bigger or smaller than the head. If it is larger, add the heads of several complements and multiply by ten, and vice versa.

For example: 46×75

(4+ 1)*7=35

6*5=30

5-7=-2

2*4=8

3530-80=3450

-

3450

3.4. The first number of a factor is one less than the last number, and the ten numbers of a factor are multiplied by two numbers whose sum is equal to 9.

Method: Add 1 to the first place of 9, and then multiply it by the first place's complement, and the number obtained is the front product. Multiply the complement of the mantissa of the first digit less than the mantissa by the number of 9 and add 1 to the back product. No ten digits complement 0.

For example: 56×36

10-6=4

3+ 1=4

5*4=20

4*4= 16

-

20 16

3.5. Two digits of different numbers in two factors are multiplied, and the tails are complementary.

Method: Determine multiplier and multiplicand, and vice versa. Multiply the multiplier head by one, the number is the front product, the tail is multiplied by the tail, and the number is the back product. Let's see if the head of the multiplicand is bigger or smaller than the head of the multiplier. If it is large, add the tails of several multipliers and multiply by ten, and vice versa.

For example: 74×56

(7+ 1)*5=40

4*6=24

7-5=2

2*6= 12

12* 10= 120

4024+ 120=4 144

-

4 144

3.6, two-factor head-tail difference one, mantissa complementary algorithm.

Method: Don't bother with the fifth one. The number obtained by subtracting one from the first square of a large number is the front product, and the hundred after rounding the tail square of a large number is the back product.

For example: 24×36

3 & gt2

3*3- 1=8

6^2=36

100-36=64

-

864

3.7. Two-digit algorithm close to 100

Method: Determine multiplier and multiplicand, and vice versa. The multiplicand subtracts the multiplier's complement to get the front product, and then multiplies the two complements to get the back product (if it is less than 10, it is filled with 0, if it is full, it is 1).

For example: 93×9 1

100-9 1=9

93-9=84

100-93=7

7*9=63

-

8463

B, square fast calculation

Find the square of 1 1 ~ 19 first.

Ibid.: 1.2. When the number of digits of the multiplier is added to the multiplicand, the number is the front product. When the digits of two numbers are multiplied, this number is the post product, full 10, the first one.

For example: 17 × 17

17 + 7 = 24-

7 × 7 = 49

-

289

3. The unit is the square of two digits of 5.

Ibid., 1.3, ten digits plus 1 multiplied by ten digits, followed by 25.

For example: 35 × 35

(3 + 1)× 3 = 12 -

25

-

1225

Four digits or ten digits are the square of five digits.

Same as above, 2.5, one digit plus 25, followed by the square of one digit.

For example: 53 ×53

25 + 3 = 28 -

3× 3 = 9

-

2809

4. Square of 2 1 ~ 50 digits

When finding the square of two numbers between 25 and 50, simply remember the square of 1~25,1~19 refer to the first article. You should remember the following four data:

2 1 × 2 1 = 44 1

22 × 22 = 484

23 × 23 = 529

24 × 24 = 576

Find the square of two digits from 25 to 50, subtract 25 from the radix, the number is the front product, and the square of the difference obtained by subtracting the radix from 50 is the back product, which is full of 1, and there is no ten digits to make up 0.

For example: 37 × 37

37 - 25 = 12 -

(50 - 37)^2 = 169

-

1369

C. addition and subtraction

First, the concept and application of complement

The concept of complement: Complement refers to the number left after subtracting a certain number from 10, 100 and 1000. ...

For example, if 10 minus 9 equals 1, then the complement of 9 is 1, and vice versa.

Application of Complement: Complement is often used in fast calculation. For example, find the multiplication or division of two numbers close to 100, and turn the seemingly complicated subtraction operation into a simple addition operation.

D, quickly calculate the division

I When a number is divided by 5,25, 125.

1, dividend ÷ 5

= dividend ÷ (10 ÷ 2)

= dividend10× 2

= dividend × 2 ÷ 10

2. Dividends ÷ 25

= dividend × 4 ÷ 100

= dividend × 2 × 2 ÷ 100

3. Dividends125

= dividend × 8 ÷ 1000

= dividend × 2× 2 ÷1000

In addition, subtraction, multiplication and division, division is the most troublesome. Even if the speed algorithm is used, it is often necessary to add pen calculation to work out the answer faster and more accurately. Because of my limited level, the above algorithm is not necessarily the best heart algorithm.

An example of fast calculation method

Examples of fast calculation in practice

○ The stone harvest speed algorithm is easy to learn and use. The algorithm starts from the high position and remembers 26 formulas summarized by the professor of history (these formulas are scientific and interrelated, and need not be memorized) to represent the carry rule of multiplying one digit by multiple digits. If you master these formulas and some specific rules, you can quickly perform operations such as addition, subtraction, multiplication, division, multiplication, root, fraction, function and logarithm.

□ This article illustrates multiplication with examples.

○ Fast algorithm, like traditional multiplication, needs to process every bit of the multiplier bit by bit. We call the number being processed in the multiplicand "standard", and the number from the first digit to the last digit on the right side of the standard is called "last digit". After the standard is multiplied, only the single digit of the product is taken as "this bit", and the number to be carried after the standard is multiplied by the multiplier is "the last bit".

○ The number of digits of the product is the number of digits of the sum of "this addition and last addition", that is-

□ Single digit of total standard = (last ten digits)

○ Then when we calculate, we should find the root and reciprocal bit by bit from left to right, and then add them to get their single digits. Give correct examples to illustrate thinking activities in calculus.

(Example) Fill in 0 before the first digit of the multiplicand and list the formula:

7536×2= 15072

The carry rule of multiplier 2 is "2 full 5 decimal 1"

7×2 original 4, later 5, full 50% 1, 4+ 1 get 5.

5×2 is 0, and if the last digit 3 is not input, it is 0.

3×2 is a 6 and the last digit is 6. When 5 is full, enter 1, 6+ 1 and get 7.

6×2 This is a 2, and there is no postposition, so you get 2.

Here are only the simplest examples for readers' reference. As for multiplication 3, 4 ... to multiplication 9, there are certain carry rules. Limited to space, I can't list them all.

Based on these carry rules, a "historical harvest fast algorithm" is gradually developed. As long as it is skillfully used, the purpose of calculating four multi-digit operations quickly and accurately can be achieved.

& gt& gt exercise example 2

□ Mastering the know-how The human brain is better than the computer.

The speed algorithm of stone harvest is not complicated, but it is easier to learn, faster and more accurate than the traditional calculation method. Professor Shi Fengshou said that ordinary people can master the tricks as long as they study hard for one month.

For accountants, businessmen and scientists, fast algorithms can improve the calculation speed and work efficiency; For students, it can develop their intelligence, use their brains flexibly, and help improve their math and physics abilities.