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: Number multiplication. If two digits are not enough, please 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: 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 = 8 2+4 = 6/kloc-0 /×1=121× 41= 8 61.
5. 1 1 Multiply by any number: Formula: the head and tail are still, and the sum in the middle is pulled down. 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. Any number more than ten times: Formula: The first digit of the second multiplier does not fall, the single digit of the first factor is multiplied by each digit after the second factor, the next digit is added, and then it falls. For example: 13×326=? Solution: 13 bit is 33× 3+2 =113× 2+6 =123× 6 =1813× 326 = 4238 Note:
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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 remains unchanged, and the number 4 with large decimal must be added with1).
Calculation method: 3×(4+ 1)= 15 (front product), 3×6= 18 (back product).
Two products 15 18.
For example, (2)84×43=36 12 (the sum of single digits is less than 10, the number 4 with small decimal places remains unchanged, and the number 8 with large decimal places adds 1).
Calculation method: 4×(8+ 1)=36 (front product) and 3×4= 12 (back product).
Adjacent combination of two products: 36 12
Such as (3)48×26= 1248
Calculation method: 4×(2+ 1)= 12 (front product) and 6×8=48 (back product).
Composition of two products: 1248
For example, (4)245 square =60025
The calculation method is 24×(24+ 1)=600 (previous product) and 5×5=25.
Composition of two products: 60025
The Weibull coefficient of ab× CD = (a-c )× d+(b+d-10 )× C.
"Head joint, tail to tail, zero as a whole, complement the remainder."
1. Find the Wei coefficient first.
2. Head-to-head multiplication (one of which adds one) is the front product (the number of adaptive tail addition is 10).
3. The tail multiplied by the tail is the back 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, the calculation method: 7×(6+ 1)=49, 3×8=24, the two products make up 4924, and then subtract 1 from the ten digits, and the final product is 49 14.
Common formulas for fast calculation (3)
(1) Multiply a dozen by a dozen.
A dozen times, a dozen times,
This method is the simplest,
Keep ten places and add one place,
Add zero and a bit product.
Prove: Let m and n be any integer between 1 and 9, then
( 10+m)( 10+n)
= 100+ 10m+ 10n+Mn
= 10〔 10+(m+n)〕+mn .
For example: 17×l6
∫ 10+(7+6)= 23 (third sentence),
∴ 230+7× 6 = 230+42 = 272 (fourth sentence),
∴ 17× 16=272。
(2) Two digits are multiplied by the same ten digits and the complement (the sum is 10).
Ten are the same, one is complementary,
Remember to multiply two numbers:
Ten plus one times ten,
The accumulation of bits goes hand in hand.
Prove: Let m and n be any integer between 1 and 9, then
( 10m+n)〔 10m+( 10-n)〕
= 100 m (m+ 1)+n( 10-n).
For example: 34×36
∫(3+ 1)×3 = 4×3 = 12 (third sentence),
The product of bits is 4x6 = 24,
∴34×36= 1224。 (fourth sentence)
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 1 1.
Two digits multiplied by 1 1,
This number is on both sides,
Leave a space in the middle,
Use and fill in.
Prove: Let m and n be any integer between 1 and 9, then
( 10m+n)×( 10+ 1)= 100m+ 10(m+n)+n .
Example: 36×ll
∵306+90=396,
∴36× 1 1=396。
Note: When the sum of two digits is greater than 10, the hundred digits will become m+ 1.
For example:
84× 1 1
∵804+ 12× 10=804+ 120=924,
∴84× 1 1=924。
Two-digit multiplication formula general formula:
The first product ranks first, and the sum of the first and last cross products is ten times the mantissa product. For example, 37x64 =1828+(3x4+7x6) x10 = 2368.
1, the same tail is complementary, the first digit is multiplied by a larger number, and the product of mantissa follows. Such as: 23×27=62 1
2. The tail is complementary to the first, the product of the first plus the tail, and the product of the mantissa follows. 87×27=2349
3. If the first digit is a mantissa complement, reduce the square of the first and last digits of a large number. For example, 76×64=4864
4, the last bit is one, the product of the first bit is followed by the sum of the first bit, followed by the product of the mantissa. Such as: 51× 21=1071.
The calculation of "several eleven times several eleven" is special: it is used for the square with the unit of 1, such as 2 1×2 1=44 1.
5, the first is different from the tail, a number plus other tails, integer multiplied by the mantissa product. 23×25=575
Quick calculation 1), the first digit is one, a number plus other tails, and 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.
Quick calculation 2) the first place is, a number plus other tails, 20 times the mantissa product. 25× 29 = 725- "Twenty times twenty"
Quick calculation 3) The first digit is 5, followed by mantissa product, and the sum of 100 digits and mantissa half. 57× 57 = 3249- "Fifty times fifty"
Quick calculation 4) If the first digit is nine, eighty plus two mantissas, followed by the product of mantissa's complement. 95× 99 = 9405- "More than ninety times more than ninety"
Quick calculation 5) The first place is four squares, fifteen with tail, followed by the square of tail. 46× 46 = 2 1 16- "40 square meters"
Quick calculation 6) The first place is the square of 5, 25 with tail, followed by the square of mantissa. 51× 51= 2601-"50 square"
6, the complement multiplied by the number of iterations, the first plus one multiplied by the number of iterations, followed by 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 certain number is multiplied by one, the head and tail are open, 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. Such as15/kloc-0 /×15 = 2265, 246× 15 =3690.
10, one hundred times one hundred, one number plus other mantissas, followed by the product of mantissas. Such as108×107 =11556.
1 1, if the difference between two numbers is 2, the square of the average of the two numbers is reduced by one. Such as 49X51= 50x50-1= 2499.
12, the number of numbers is multiplied by the number of nine, this number is subtracted from the difference of (the number of the first few numbers+1) to make the first few numbers of the product, and the last number and the unit make up a few 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.
2) Multiply a number by 99: this number is subtracted (the first ten digits+1) and the last two digits add up to100:14× 99 =14-(0+1) =/kloc-. 100- 14=86 1386 158×99= 158-( 1+ 1)= 156, 100-58=42 15642 7357×99= 7357-(73+ 1)=7283 100-57=43 728343
3) Multiplying a number by 999: It can be inferred according to the above method: this number is subtracted (the first few hundred digits+1), and the last three digits add up to10001234× 999 =165438-(6234.