First, add a method-the head is unchanged, and the sum of digits is equal to 10.
Formula: after adding "1" to a person's head, the person's head × the person's head; Tail × tail, connection.
For example: 62×68=42 16.
Solution: (6+ 1)×6=42 2×8= 16, which are connected to form 42 16.
Exercise: 73×77 28×22 64×66 43×47.
Second, the mantissa method-mantissa addition, the sum of ten digits is equal to 10.
Formula: head × head plus a tail; Tail to tail
For example: 26×86=2236
Solution: 2×8+6=22 6×6=36 is connected to get 2236.
Exercise: 38×78 47×67 85×25 64×44.
Third, the method of subtracting 1-the single digit is 1 and the difference between the two prefixes is 1.
Formula: subtract 1 from the square of the first digit of the larger number, and then 99.
For example: 8 1 (large number) ×79=6399.
Solution: 82- 1=63 followed by 99 to get 6399.
Exercise: 6/kloc-0 /× 59 7/kloc-0 /× 69 29× 3149× 51.
Fourth, find the product of two hundred digits, and the mantissa method of adding one number to the other.
Formula: one number+the mantissa of another number; Tail × tail, connection.
For example:105×107 =11235.
Solution:105+7 =125× 7 = 35 to get 1 1235.
Exercise:108×109106×104102×108103×105.
5. 1, find the square number of 5 1-59, and add the mantissa method to the constant. (Constant is 25)
Formula: constant 25+ tail; Tail × tail, connection.
Example 1, 582=3364 solution: 25+8=33 8×8=64, and 3364 is obtained by connecting them.
Example 2, 532=2809 Solution: 25+3=28 3×3=09, connected together to get 2809.
Exercise: 542 562 572 522
2. Find the square number of 4 1-49, and the complement method of subtracting single digits from constants.
Fill the single digits with 10, and you can find the complement of the single digits. If the complement of bit 4 is 6, the complement of 6 is 4 and the complement of 2 is 8.
Formula: the complement of constant 25 minus single digit; Complement × complement, connection.
For example, 1, 462=2 1 16.
Solution: The 6' s complement is 4,25-4 = 21.4× 4 =16, and the concatenation results in 2 1 16.
Example 2,482 = 2304
Solution: The complement of the eighth bit is 2,25-2 = 23 2× 2 = 04, and the result is 2304.
Exercise: 472 482 452 492
3. Find the square of a number with 5 digits.
Formula: head+1× head; Tail × tail together.
Example: 852=7225
Solution: (8+ 1)×8=72 5×5=25 to get 7225.
Exercise: 352 652 752 452
4. Find the square of 9 1-99; The complement method of subtracting single digits from this number.
Formula: this number MINUS the complement of single digits; Complement × complement, connection
Example 1, 942=8836
Solution: 94-6=88 6×6=36, and 8836 is obtained by connection.
Example 2,982 = 9604
Solution: 98-2=96 2×2=04, followed by 9604.
Exercise: 952 972 962 992
6. Find the product of any number and 1 1.
For example: 1, 235×11= 2585748×11= 8228.
2 3 5 7 4 8
2 5 8 5 7 1 1 12 8
Methods: Write from the beginning to the end, write the composite number in the middle, and enter ten into one.
Exercise: 816×114536×19247×115672×1.
Multiply 7999 by any number
Formula: After subtracting "1" from the end of any number, write its parity complement.
What is complement? A number that can change one digit into 10, two digits into 100 and three digits into 1000 is called a complement.
For example, the complement of 7 is 3, the complement of 42 is 58, and the complement of 472 is 528.
For example, 1, 999×5 16=5 15484.
Solution: The complement of 516-1= 515516 is 484, and the conjunction is 5 15484.
Example 2,999× 74 = 73926
Solution: The parity complement of 74- 1=73 074 is 936, written as 73926.
Exercise: 999×547 999×873 999×67 999×82.
Multiply 999 by multiple numbers:
999×2437=2434563
Solution: 2437-(2+ 1)=2434, the complement of apposition 437 =563, and the link is 2434563.
999×24738=247 13262
Solution: 24738-(24+1) = 24713, the complement of apposition 738 =262, and the conjunction is 247 13262.
Exercise: 999× 3576 999× 5628 999× 24736 999× 51472.
Eight, universal method-arbitrary number multiplication (three examples can be applied only after they are fully understood).
Formula: square of internal and external terms, product addition, head × head+head; Tail × tail ten plus tail together.
For example, 1, 62×57=3534.
Solution: ○ 1 square of internal and external terms, product addition.
2 (internal item) ×5 (internal item) = 10 6 (external item) ×7 (external item) =42
10+42=52
○2 First memorize the sum of the inner and outer products "52", then add the first 5,6× 5+5 = 35 of "52" to the head× head, and add the mantissa 2,2× 7 =14 to form 34 conjunctions.
Exercise: 43×58 23×46 72×85 93×64.
Example 2, 63*82=5 166
Solution: ○ 1 square of internal and external terms, product plus 3×8+6×2=36.
○2 Memorize the sum of inner and outer products of 36, then add the first 3,6× 8+3 = 51of "36" to the head× head, and add the mantissa 6,3× 2 = 06 to the tail× ten digits, and add 6 to get 66, which is recorded as 5 166.
Exercise: 74× 625198 83× 53 82× 73.
Example 3, 38+56 = 2 128
Solution: ○ 1 square of internal and external terms, product plus 8×5+3×6=58.
○2 Recite "58" first, then: head × head plus the first 5 of "58", 3×5+5=20, tail× tail plus the mantissa of "58", 8× 6 = 48, and decimal digits plus 8, so as to get 128 20 and1.
Exercise: 47×69 74×38 89×35 56×68
Attachment: Fast test algorithm of multiplication and division-discard 9 remainder test algorithm.
In this way, you don't need to write, save time and brain, and it's quick and clear at a glance.
1. What does it mean to discard the remainder of 9?
Discard a number with 9 digits or add it arbitrarily to get 9, add the remaining digits, the added digits will be greater than 9, and then add the obtained digits until it is less than 9. For example:
32966472 Discard its 9 first, then discard its 3 plus 6 to get 9, 2 plus 7 to get 9, and add the remaining 6 plus 4 plus 2, 6+4+2= 12, 12 is greater than 9, and then add up, 1+2=3.3 is less than 9.
2. Multiplication discarding nine test algorithm: visually calculate the remainder of each number on both sides of the equal sign. If both sides are equal, the calculation is correct; if not, the calculation is wrong.
For example: 5349×746=3990354. Check whether the calculation is correct by discarding the remainder of 9.
Left inspection: 5349× 746 3 (7+4+6) 3×173× (1+7) 3× 8 242+4 = 6.
Number on the right: 3990354 3+3=6.
Left 6= right 6 both sides are equal, and the calculation is correct.
(In the practical application of the fast method of checking the remainder of discarding 9, the whole process is visually calculated orally, without pen calculation, and the eyes and hearts are consistent and can be combined. If the sum of several numbers is 2-3 times of 9, it can also be discarded. )
3. Divide and discard 9 test algorithm: Divider and discard 9 remainder = Divider and discard 9 remainder × Quotient and discard 9 remainder (the method is the same as multiplication).
Try to discard the 9 remainder test algorithm and check whether the following questions are calculated correctly.
4252×6 13=2606476 4359×86 1=3752099
6 137× 145=889865 63885 15÷765=835 1
5604 152÷365= 15742 3265866÷92 1=3546
(2) Quick-acting No.2 prescription
First of all, add some reasons:
Where the single digit of the number to be squared is 1, and this number is greater than the power of 10 or the multiple of 10, adding the multiple of 10 to the last digit of 10 is the root of this number.
For example: =1110 =100 <121.
10+ 1= 1 1
=5 1 50×50=2500 84 1 30- 1 = 29。
=39 40×40= 1600> 152 1 40- 1=39
=99 100× 100= 10000>980 1 100- 1=99
Three plus five theorem:
The number of units of a square is 5. When this number is greater than the power of 10 or the multiple of 10, the root of this number is 10 or the multiple of 10 plus the last digit of 5.
For example: = 25 20× 20 = 400 < 625 20+5 = 25.
=65 60×60=3600