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: Numbers are multiplied. 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 = 621Note: Multiply the digits. If there are not enough digits, use 0 to fill 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= 44× 4 =167× 4 = 28 37× 44 =1628 Note: Number multiplication. If two digits are not enough, please use 0. 4. Dozens of eleven times dozens of eleven: formula: head to head, head to head, tail to tail. For example: 2 1×4 1=? Solution: 2× 4 = 82+4 = 6/kloc-0 /×1=121× 41= 8615.1Multiply any number: formula. For example: 1 1×23 125=? Solution: 2+3 = 53+1= 41+2 = 32+5 = 72 and 5 are at the beginning and end, respectively.1× 23125 = 254375 Note: If it is full, 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: I saw the "one-minute fast formula" on TV and thought it was very good, so I wanted to share it with you: when the ten digits are the same and the sum of the single digits is equal to 10, for example, 62×68=42 16.
Calculation method: 6×(6+ 1)=42 (front product) and 2×8= 16 (back product). The theorem of the special problem in the one-minute quick calculation formula is: the product obtained by multiplying any two digits by any two digits, as long as the Wei coefficient is "0", must be the product obtained by multiplying the tail in two binomials, and the product obtained by multiplying the head by the head (the sum of one term plus 1) is the product of the previous term and the product obtained by two adjacent products. For example, (1) 33× 46 =1518 (the sum of single digits is less than 10, so the number 3 with small decimal places remains unchanged, and the number 4 with large decimal places must be added with1). Calculation method: 3× (4+ 1). The product of 3×6= 18 (after product) constitutes 15 18, such as (2)84×43=36 12 (the sum of single digits is less than 10, and the number 4 with small decimal places remains unchanged. 3×4= 12 (back product) Two adjacent products: 36 12, such as (3)48×26= 1248 Calculation method: 4× (2+1) =1. 2. The first product (one of which is plus one) is the first product (the number that adapts to the tail addition is 10). The last product is the post-product. 4. The two products are connected, and the Wei coefficient can be added to ten digits. For example, 76×75 and 87×84, where the same ten digits add up to 1 1, then its Weibull coefficient must be these ten digits. For example, the 76×75 Weibull coefficient is 7, and the 87×84 Weibull coefficient is 8. For example: 78×63, 59×42, their coefficients must be ten digits minus its single digits. For example, the Weibull coefficient of the first question is equal to 7-8=- 1, and that of the second question is equal to 5-9=-4. As long as the ten digits differ by one digit, you can quickly calculate the number whose single digits add up to 1 1 by the above method. Example: 1 76×75, the calculation method: (7+ 1)×7=56 5×6=30, two products make up 5630, and then add 7 to the decimal number, and the final product is 5700. Example 2 78×63, calculation method: 7×(6+ 1)=49, 3×8=24, the two products add up to 4924, and then subtract 1 from the ten digits, and the final product is 49 14 (three commonly used fast formulas) (/kloc It is proved that if m and n are arbitrary integers from/kloc-0 to 9, Then (10+m) (10+n) =100+10n+Mn =10 [Example:1 It is proved that if m and n are arbitrary integers from 1 to 9, then (10m+n) [10m+(10-n)] =100m (m+1)+ Example: 34× 36 ∶ (3+1)× 3 = 4× 3 =12 (third sentence), and the product of units is 4× 6 = 24, ∴ 34× 36 = 1224. Note: When the product of two numbers is less than 10, the decimal number should be written with zero. (3) Multiply any other two digits by eleven with 1 1, go to both sides of this number, leave a space in the middle, and use and make up. It is proved that if m and n are arbitrary integers from 1 to 9, then (10m+n) × (10+1) =100m+10 (m+n)+n. Note: When the sum of two digits is greater than 10, the hundred digits will become m+ 1, such as 84×1:804+12×10 = 804+. 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=1071-It is special to quickly calculate "several eleven times several eleven": it is used for the square with the unit of1,such as 2/kloc-0 /× 2/kloc. 23×25=575 (1), the first digit is the same, and a number plus other mantissas is the product of ten mantissas. 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" 6. If the complement is multiplied by the number of iterations, the first one is multiplied by the number of iterations, and then the product of 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" 8. If a number is multiplied by one, the head and tail are pulled apart, and the sum of the head and tail stands 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 =