2. Complementarity of the same head and tail (the sum of the tails is equal to 10): Complementarity of the same head and tail (the sum of the tails is equal to 10): formula: one head plus one formula: after one head is added, the head is multiplied by the head and the tail is multiplied by the tail. Head joint, tail-to-tail example: 23×27=? 23×27=? Solution: 2+1= 32× 3 = 63× 7 = 2123× 27 = 621. Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space. Unit multiplication, 3. The first multiplier is complementary, and the other multiplier has the same number: the first multiplier is complementary, and the other multiplier has the same number: formula: one head plus one formula: after one head is added, the head is multiplied by the head and the tail is multiplied by the tail. Head joint, tail to tail example: 37×44=? 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. Unit multiplication, 4. Dozens of eleven times dozens of eleven: dozens of eleven times dozens of eleven: formula: head to head, head to head, tail to tail. Formula: head joint, head joint, tail to tail. For example: 2 1×4 1=? 2 1×4 1=? Solution: 2× 4 = 8 2+4 = 6/kloc-0 /×1=121= 861times any number: 5.1. Formula: head and tail do not move down, middle and pull down. For example: 1 1×23 125=? 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, If you add up ten, you will get one. 6. Multiply a dozen by any number: multiply a dozen by any number: formula: the first digit of the second multiplier does not fall, formula: the first digit of the second multiplier does not fall, the unit of the first factor multiplies each digit after the second factor, and then falls. Count, add the next number, and then drop.
For example: 13×326=? 13×326=? Solution: 13 bit is 33× 3+2 =113× 2+6 =123× 6 =1813× 326 = 4238 Note: If you add up ten, you will get one.