I graduated from junior high school. I don't think I'm very good at math, so it's difficult to calculate double digits orally. It took me a long time to figure out, is there any way for me to learn ma

A method to calculate two numbers quickly. The sum of one of these two numbers is 10, and the other numbers are the same. Add 1 to the front of the number, and add the product of the number to the end of the number. For example: 56× 545+ 1 = 6, 6× 5 = 30. Add the product of 4 and 6 at the end of 30 to get 3024, so 56× 54 = 3024. Another example: 6 1 × 69 (6+ 1) × 6 = 42, 1 × 9 = 9. When the product of the number in a unit is a single digit, it still occupies two digits, so a 0 should be added before 9. So 6 1× 69 = 4209.

Second, the same ten digits, the sum of single digits is not 10, and a fast calculation method for multiplying two digits.

Add a number in the unit of another number, multiply it by an integer consisting of numbers in the tenth place, and add the product of two numbers in the second place. For example: 53× 54 = (53+4 )× 50+3× 4 = 57× 50+12 = 2850+12 = 2862.

3. A fast calculation method of multiplying the sum of the digits in the tenth digit by the two digits of 10 when the digits in one digit are the same. Multiply ten digits by one digit, and finally add one digit product. (If the unit product is less than two digits, add 0 before the product to make up the two digits), for example, 24×84 times ten digits and one digit: 2× 8+4 = 20, and the unit product is 4× 4 = 16, so 24× 84 = 20 16. Exercise: 35×75 17×97 48×68

Four, the sum of each digit is 10, and the fast calculation method of multiplication with two digits with the same number of digits.

Add the product of the number and the two digits of 10 plus 1 multiplied by the ten digits of the same two digits to the end of the product. (If the product of digits is less than two digits, add 0 to make up the two digits) For example, the product of 46×33 digits and the ten digits of 1 multiplied by the ten digits of the same two digits: (4+ 1) × 3 = 15, and the product of single digits is 3× 6 =.

Five: A fast calculation method for the difference between the sum of two digits multiplied by 10 and the digits of 1. The number is multiplied by the square of 10 minus the square of the decimal place. Such as: 46× 34 = (4×10 )× (4×10)-6× 6 =1600-36 =1564.

1 .10 times 10: formula: head joint, tail to tail, tail to tail. For example: 12× 14=? Solution:/kloc-0 /×1=12+4 = 62× 4 = 812×14 =168. Note: Multiply by single digits. If two digits are not enough, please use 0. 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= 32× 3 = 63× 7 = 2123× 27 = 621. Note: Multiply the numbers. If two digits are not enough, use 0. 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= 44× 4 =167× 4 = 28 37× 44 =1628.

Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space. 4. Dozens of eleven times dozens of eleven: formula: head to head, head to head, tail to tail. For example: 2 1×4 1=? Answer: 2×4=8

2+4 = 61×1=121× 41= 8615.1times any number: formula: 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,11× 23125 = 254375 Note: If you add ten, you must enter one. 6. Multiply a dozen by any number: Formula: The first digit of the second multiplier does not move down, the single digit of the first factor is multiplied by each digit after the second factor, plus the next digit, and then falls down. For example: 13×326=? Solution: 13 bit is 33× 3+2 =113× 2+6 =123× 6 =1813× 326 = 4238 Note:

The general formula for fast calculation of two-digit multiplication is: the product of the first digit ranks first, and the sum of the cross products of the first digit and the last digit is ten times the mantissa product. For example, 37x64 =1828+(3x4+7x6) x10 = 23681,the tails are complementary, the first digit is multiplied by a larger number, and the product of mantissa follows. For example: 23×27=62 1 2, the tail is complementary to the head, the product of the first digit is added to the tail, and the product of the mantissa is added to the tail. 87×27=2349 3. If the mantissa is complementary to the first one, reduce the square of the beginning and end of a large number. For example, 76×64=4864 4, where the last bit is the same, the product of the first bit is followed by the sum of the first bits, followed by the product of the mantissa. For example: 5 1 × 21=1-the calculation of "several eleven times several eleven" is quite special: it is used for the square of the unit1,such as 21× 2/kloc-0. 17× 19 = 323—— The fast calculation of "ten times ten" includes that the ten digits are the square of 1 (i.e.1~19), such as1. 25×29 = 725—— "20 times 20" quick calculation 3) The first digit is five, followed by the mantissa product, and the sum of hundreds and mantissa half. 57× 57 = 3249-"50 times 50" quick calculation 4) If the first digit is nine, eighty plus two mantissas, followed by the product of mantissa's complement. 95 × 99 = 9405 —— "90 odd times 90 odd" quick calculation 5) The first one is the square of Siping, with fifteen tails, followed by the square of the tail. 46× 46 = 2116—"40 square" quick calculation 6) The first one is five squares, 25 follows the tail, and the mantissa square follows. 51× 51= 2601-"50 square" quick calculation 6. If the complement is multiplied by the iteration number, the first digit is multiplied by the iteration number, and then the product of the mantissa. 37×99=3663 7. If the last digit is five squares, the first digit is multiplied by one and then multiplied by the product of mantissa. For example, 65× 65 = 4225-"the square of several fifteen" is a quick calculation. 8. If a certain number is multiplied by 1, the head and tail are open and the sum of the head and tail is in the middle. For example, 34×11= 3 3+4 4 = 374 9, a number is multiplied by 15, and the original number is added with half of the original number and then followed by a 0 (the original number is even) or moved one place after the decimal point. For example,15/kloc-0 /×15 = 2265, 246× 15 =3690 10, one hundred times one hundred, one number plus another mantissa, and the product of mantissa follows. If108×107 =11551,and the difference between the two numbers is 2, the square of the average of the two numbers is reduced by one. For example, 49X51= 50x50-1= 249912, the number of digits is multiplied by the number of 9, and the difference between the first few digits (+1) is subtracted from this number to make the first few digits of the product, and the last digit and the single digit form several zeros. 1) Multiply a number by 9: this number subtracts the first few digits of the product (the first few digits of a unit+1), and the last digit and the unit complement 10 4×9=36. Think about it: 0,4-(0+1) = 3 in front of the unit, and the last one is. 783-(78+ 1) = 704, and the last digit is 10-3 = 7) Multiply a number by 99: subtract this number (the first ten digits are+1), and the last two digits add up to100: 650. 100- 14 = 86 1386 158×99 = 158-( 1+ 1)= 156， 100-58 = 4215642 7357× 99 = 7357-(the last three digits of 73+65438 add up to10001kloc-0/234× 999 =

1.9 (two-digit multiplication) The formula of mathematical quick calculation skill 9:1× 9 = 92× 9 =183× 9 = 274× 9 = 36, 5× 9 = 456× 9 = 547× 9 = 638. In the first grade of primary school, you may only learn addition, but in the first semester of the second grade, you will learn multiplication formula. In fact, many parents may have taught the above formula when their children are not in school. But did the children take a closer look at the characteristics of the formula above? From the above formula, what you see is that any number from 1 to 9 is multiplied by 9, and the sum of single digits and ten digits is still equal to 9. Look at the top: 0+9 = 9; 1 + 8 = 9; 2 + 7 = 9; 3 + 6 = 9; ,4 + 5 = 9; 5 + 4 = 9; 6 + 3 = 9; 7 + 2 = 9; 8+ 1 = 9, maybe children will ask, what's the use of discovering this secret? My answer is very useful. This is the basis for training you to be good at observing, summarizing and finding out the laws of things. Let's do some more complicated multiplication: 18 × 12 =? 27 × 12 = ? 36 × 12 = ? 45 × 12 = ? 54 × 12 = ? 63 × 12 = ? 72 × 12 = ? 8 1 × 12 = ? The multiplication of two digits may not be learned until the third grade, but did the children see that in the above topic, the previous multipliers are all multiples of 9, and the sum of one digit and ten digits is equal to 9? So can we find a simple algorithm? That is to say, turn the multiplication of two digits into the multiplication of one digit? Let's change these figures first. 18 = 1 × 10 + 8; 27 = 2 × 10 + 7; 36 = 3 × 10 + 6; 45 = 4 × 10 + 5; 54 = 5 × 10 + 4; 63 = 6 × 10 + 3; 72 = 7 × 10 + 2; 8 1 = 8 × 10 + 1; Shall we change the numbers on it again? 1×10+8 =1× 9+1+8 =1× 9+9 =1× 9+2 × 9. Of course, if you know the formula, you can put 6544 directly. Similarly, you can break down the following numbers or recite formulas. Go back and practice by yourself. 27 = 3 × 9 ; 36 = 4 × 9 ; 45 = 5 × 9 ,54 = 6 × 9 ; 63 = 7 × 9 ; 72 = 8× 9,8 1 = 9× 9. In order to find a method to calculate the above problem, we change the above formula again. 18 = 2×( 10- 1); 27 = 3×( 10- 1); 36 = 4×( 10- 1)45 = 5×( 10- 1); 54 = 6×( 10- 1); 63 = 7×( 10- 1),72 = 8×( 10- 1); 81= 9× (10-1). Now let's calculate the above question:18×12 = 2× (10-1). = 2 ×( 120- 12), the children in brackets should know addition, that is, they will know it in the first grade. 120 - 12 = 108; So there it is,18×12 = 2×108 = 216. Did you change the multiplication of two digits into the multiplication of one digit? And you can get the result by oral calculation? Can children try it on their own? That's how I teach multiplication. He only needs me to do this one, and he will solve the later problems himself. It seems that our above calculation is very troublesome. In fact, it is very simple to sum up now. Look at the next topic: 27×12 = 3× (10-1)×12 = 3× (120-12) = 3×/kloc.

36×12 = 4× (10/)×12 = 4× (120-12) = 4×100. The following questions seem unnecessary. It's all about adding the previous number 1 and multiplying it by 108, 45 × 12 = 5 × 108 = 540, 54×12 = 6×108 = 60.

72×12 = 8×108 = 864, 81×12 = 9×108 = 972. Let's take a look at the above calculation results. Did the children find anything? We changed a two-digit multiplication into a one-digit multiplication. The sum of one digit and ten digits of a multiplier is equal to 9, so that the multiplier of one digit in the changed number is just larger than the previous multiplier 1. Another feature of the last two digits is that it is a hyphen (12), and 1 and 2 are continuous. Can you find a simpler calculation method? In order to find a simpler algorithm. Here I introduce a new term-children's complement. What is complement? Because this term is simple, even kindergarten children will understand it quickly. , 1 + 9 = 10; 2 + 8 = 10; 3 + 7 = 10; 4 + 6 = 10; 5 + 5 = 10; 6 + 4 = 10; 7 + 3 = 10; 8 + 2 = 10; 9 + 1 = 10; As can be seen from the above addition, if the sum of two numbers is equal to 10, then the two numbers are complementary. That is to say, 1 and 9 are complements, 2 and 8 are complements, 3 and 7 are complements, and 4 and 6 are complements. You don't need to remember the complement of 5 or 5, just remember 4. Now let's take a look at the above calculation results: take a 63 × 12 = 7 × 108 = 756 as an example. The first number of the result is 7 (regardless of its position). Is it exactly equal to the first digit in the first multiplier (63) plus 1? 6+ 1 = 7, how to calculate the last two digits of the result? If this 7 is multiplied by the last complement (8) of the following multiplier (12), what will it be? 7 × 8 = 56 Hehe, we don't need to decompose now, we just need to add 1 to the first number of the result, and then multiply this number by the last complement (8) of the next multiplier (12) to get the last two digits of the result. Is this ok? If it works, it's really too fast. It is really a quick calculation.

Try other questions: 18 × 12 = the number before the first multiplier (18) plus 1: 1+ 1 = 2- the number before the result.

The complement of the number (2) after multiplying 2 by the second multiplier (12) (8): 2× 8 = 16.

The result is 2 16. Take a look at your face? 27 × 12 =, and the first number in the result is -2+ 1 = 3.

The last number of the result is -3× 8 = 24, the result is 324, 36 × 12 =, and the first number of the result is -3+ 1 = 4.

The last number of the result is -4× 8 = 32, the result is 432,45×12 =, and the first number of the result is -4+ 1 = 5.

The last number in the result is -5× 8 = 40, the result is 540,54×12 =, and the first number in the result is -5+ 1 = 6.

The last number of results is -6× 8 = 48, the results are 648, 63 × 12 =, and the first number of results is -6+ 1 = 7.

The last number of the result is -7× 8 = 56, the result is 756,72×12 =, and the first number of the result is -7+ 1 = 8.

The last number of the result is -8× 8 = 64, the result is 864,8/kloc-0 /×12 =, and the first number of the result is -8+ 1 = 9.

The last number of the result -9× 8 = 72, the result is 972. Is the calculation result the same as the above method?

What else can the child see from the results? Is it the sum of the three digits of the calculation result or equal to 9 or multiple of 9?

Do the math yourself, don't you? Children who read this article, I will ask you a few questions to see if you have mastered the method.

54 × 34 = ? 18 × 78 = ? 36 × 56 = ? ,72 × 89 = ? 45 × 67 = ? 27 × 45 = ? 8 1 × 23 = ? Through this topic, I mainly want children to draw inferences from a topic and find some regular things from it. In this way, you can quickly master the addition, subtraction, multiplication, division and Divison of mathematics without doing too many problems. If the above topic is expanded, the following hyphen will be expanded into multiple digits. For example: 123, 234, 345, 2345, 34567, 123456, 23456789 and so on. See if there are any algorithms, maybe we can find a quick calculation method.

If possible, say 63 × 2345678 =

Such a topic can be quickly calculated by oral calculation. I believe that every child is a genius as long as he constantly summarizes scientific methods! If you can't find a way, I'll help you find a quick way tomorrow.

One, the two digits are the same, and the sum of the two mantissas is the two-digit multiplication of 10 (the first digit of the multiplicand is added by 1), and then the two digits are multiplied to get a product, and the two mantissas are multiplied to get a product, and the two products are connected to get the product. For example: 72 63 84

× 78 × 67 × 86,56 16 422 1 7224

Note: If the square mantissa of two digits is 5, this method can also be used. For example, 25× 25 = 625 45× 45 = 202575× 75 = 5625 95× 95 = 9025.

Two, two digits are the same, and the sum of two mantissas is not equal to 10. First multiply two mantissas to get a product, then multiply the sum of two mantissas by the first bit of the multiplicand to get a product, and finally multiply two numbers (the square of the first bit) to get a product, and then add three products to get the desired product. For example, 52 6 1 73, × 53 × 62 × 74, 2756 3782 5402. Note: If the square mantissa of two digits is not 5, this method can also be used. Such as: 22 66, × 22 × 66, 484 4356.

3. The first and the last of the multiplicand are the same. The sum of the first and the last of the multipliers is the two-digit multiplication of 10: (the first digit of the multiplier is added to 1), then the two mantissas are multiplied to get a product, and the two digits are multiplied to get a product, and the last two products are connected to get a product. Such as: 22 44 88, × 19 × 28 × 37.

4 18 1232 3256

4. The sum of the first two digits is 10, and two digits with the same mantissa are multiplied to get a product. The product of the first two digits, plus the same mantissa, gets a product, and these two products are the products sought. Such as: 26 76 47, × 86 × 35 × 67, 2236 2656 3 149.

5. The difference between two digits is 1, and the sum of two mantissas is 10. For example, 38×22=836 can be decomposed into (30+8)×(30-8)=30×30-8×8=836, and the principle is A× A-B.

6. Multiplication of any two digits: (cross multiplication or diagonal multiplication) First, multiply the sum of cross multiplication (the product of the first digit of the multiplicand multiplied by the mantissa of the multiplier plus the product of the mantissa of the multiplicand multiplied by the first digit of the multiplier) with the product of two first digits and two mantissas. Such as: 43×85=3655

Seven, three-digit multiplication, the first and middle numbers are the same, and the sum of mantissas is equal to 10. First, two mantissas are multiplied to get a product, (multiplied by 1) and then two medians are multiplied to get a product. Then the two medians are added and multiplied by the first bit of the multiplicand to get a product, and the last two first bits are multiplied to get a product, and the four products are connected to get a product. 1 12× 1 18 = 132 16, 1 12× 1 18, 13265448. For example,12468×1=137148,25124×1.

Multiply by 9, 9, 99, 999 and so on. Use any number: first find the complement of any number (the sum of two numbers is 10, and the two numbers are complementary), and then connect the complement to the last digit of 9, 99, 999, etc. Finally, subtract the complement from the highest digit of the new number, which is the product of the sum. Such as: 999×999=99800 1

9999×8997=8996 1003

Other people's

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. 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

14× 19=?

1× 1= 1

4+9= 13

4×9=36

14× 19=266

Note: the head is multiplied by the head, and the head is added. If two digits are not enough, use 0 for ten digits.

3. 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: 13×326=?

Solution: 13 bit is 3.

3×3+2= 1 1

3×2+6= 12

3×6= 18

The first number of 326 is 4.

13×326=4238

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

4. 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