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for ease of us to understand the universal law of double-digit multiplication, there will be through specific examples. By comparing the results of a large number of double-digit multiplication, I double-digit multiplication result is divided into three parts, bits, ten, ten more that one hundred and thousand. (Maximum two-digit multiplication is not more than 10,000, so only up to one thousand) is an example: 42 × 56 = 2352

which,tods shoes, thanks to a number of digits to determine the method is to take the number two bit product mantissa is thanks to a number of digits. Specific to the above example, 2 × 6 = 12, which,mbt footwear, thanks to a 2 mantissa, 1 bit binary number;
thanks to a ten-digit is determined, taking the number two and ten bits are cross-multiply and add a bit into the digit sum of the mantissa, as thanks to a ten digit. Specific to the above example, 2 × 5 +4 × 6 +1 = 35, where 5 is thanks to a ten-digit, 3 for the ten decimal;
thanks to a method for determining the rest is to take the number two the product of ten and ten-digit decimal number and that is thanks to a hundred or a thousand. Specific to the above example, 4 × 5 +3 = 23. It is thanks to a 2 and 3, respectively,mbt sale, of the thousands and hundreds digits.

Therefore, 42 × 56 = 2352. As another example, 82 × 97, in accordance with the above calculation method, first determine thanks to a single digit, 2 × 7 = 14, then thanks to a number of bits should be 4; then determined thanks to a ten-digit, 2 × 9 +8 × 7 +1 = 75, then thanks to a median of ten 5; calculated thanks to a final rest, 8 × 9 +7 = 79, therefore, 82 × 97 = 7954. Similarly, using this algorithm, it is easy to see that all two-digit multiplication plot.
one digit by digit.
1. dozen by dozen:
formulas: the head by head, tail plus tail, tail by tail.
Example: 12 × 14 =?
solution: 1 × 1 = 1
2 +4 = 6
2 × 4 = 8
12 × 14 = 168
Note: a bit multiply, not to use the 0 digit placeholder .

2. head the same, complementary tail (tail sum equal to 10):
formulas: a first add 1 to the head by head, tail by tail.
Example: 23 × 27 =?
solution: 2 +1 = 3
2 × 3 = 6
3 × 7 = 21
23 × 27 = 621
Note: a bit multiply, not to use the 0 digit placeholder .

3. the first multiplier complementary, the same as another multiplier:
formulas: a first add 1 to the head by head, tail by tail.
Example: 37 × 44 =?
solution: 3 +1 = 4
4 × 4 = 16
7 × 4 = 28
37 × 44 = 1628
Note: a bit multiply , not use the 0 digit placeholder.

4. dozens of dozens of one by one:
formulas: the head by head, head plus head, tail by tail.
Example: 21 × 41 =?
solution: 2 × 4 = 8
2 +4 = 6
1 × 1 = 1
21 × 41 = 861

5. 11 by any number:
formulas: the whereabouts of the fixed end to end, in the middle of the drop-down.
Example: 11 × 23125 =?
solution: 2 +3 = 5
3 +1 = 4
1 +2 = 3
2 +5 = 7
; 2 and 5, respectively, in the beginning and end
11 × 23125 = 254375
Note: and over 10 to enter one.

6. dozen by any number:
formulas: the whereabouts of the second movement is not the first multiplier, multiplying the first factor a bit behind the second factor of each number, plus the next digit, falling again.
Example: 13 × 326 =?
solution: 13 bits is 3
3 × 3 +2 = 11
3 × 2 +6 = 12
3 × 6 = 18
13 × 326 = 4238
Note: and over 10 to enter one.

double-digit multiplication math on the The so-called 3 is exactly equal to the sum of 10, I told him, multiplied by such figures as, in fact, there is a pattern. Is the two-digit number of the product as the latter thanks to a double-digit, less than 10, up on the ten digit 0; the same number of two to take one of the ten plus 1 multiplied, the result is thanks to a thousand and hundreds. Specific to the example above, 67 × 63,7 × 3 = 21, this is thanks to a number of 21 after two; 6 × (6 +1) = 6 × 7 = 42, this is thanks to a 42 the first two comprehensive up, 67 × 63 = 4221. Similarly, 15 × 15 = 225,89 × 81 = 7209,64 × 66 = 4224,Tods sale,92 × 98 = 9016. I told him this quick calculation small In the I told him, the so-called five, ten-digit result is exactly equal to 4 +6 10. Its algorithms is the number two of you the same number of digits after the product is thanks to a number of less than 10, fill in the ten on 0; two decades with the same number of bits multiplied digits, The result is that thanks to a hundred and thousand. Specific to the above example, 45 × 65,5 × 5 = 25, this is thanks to a post 25 double-digit, 4 × 6 +5 = 29, this is thanks to a front portion 29,tods sale, thus, 45 × 65 = 2925. Similarly, 11 × 91 = 1001,83 × 23 = 1909,74 × 34 = 2516,97 × 17 = 1649. topics related articles：


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