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.
This law can be deduced by algebraic method.
( 10+a)( 10+b)= 100+ 10(a+b)+ab
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 = 3
2×3=6
3×7=2 1
23×27=62 1
Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space.
Algebraic derivation of this law
( 10a+b)( 10a+c)= 100 aa+ 10a(b+c)+BC = 10a+ 10a 10+BC = 100 a(a+ 1)+BC
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=4
4×4= 16
7×4=28
37×44= 1628
Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space.
4. 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
5. 1 1 times any number:
Formula: head and tail do not move down, middle and pull down.
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.
1 1×23 125=254375
Note: If you add up to ten, you will get one.
6. 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
13×326=4238
Note: If you add up to ten, you will get one.