The calculation principle of combining the first and last digits is as follows:
The last two digits of the answer are multiplied by the last digits: the first digit is multiplied by (first + 1), and the first to two digits are taken up. bit, you can get the product.
For example: 21*29. The first number 2 of the two factors is the same, and the second number 1+9=10, so it is said that the first number and the last number are the same.
Multiply the mantissas: 1*9=9, 2*(2+1)=6, and the order is 609.
For two two-digit numbers, if the ones digits are the same and the sum of the tens digits is 10, then these two numbers are called "two-digit numbers with the same tail and the same head".
For example: 34 and 74, 45 and 65.
The multiplication of two two-digit numbers in each group can be quickly calculated as follows: (head 1 × head 2 + tail) × 100 + tail × tail.
Example 1: 34×74=(3×7+4)×104×4=2516.
Example 2: 45×65=(4×6+5)×100 +5×5=2925.
As can be seen from Example 1 and Example 2, the multiplication of two two-digit numbers with the same tail and the same head can be quickly calculated as follows:
Use "head 1 × head 2 + tail" The sum of is the first half of the product, and the product of "tail × tail" is used as the last two digits of the product.
If the product of "tail × tail" is less than two digits, add 0 to the tens place.
In addition, a tens digit is a two-digit multiplication of 5, which can also be regarded as a quick calculation of "the tail is the same as the head and the ten are joined together".
For two two-digit numbers, if the ten digits are the same and the sum of the single digits is 10, the two numbers are called "two-digit numbers with the same beginning and the end".
For example: 23 and 27, 62 and 68.
The two-digit multiplication of "the head and the tail plus ten" can be quickly calculated as follows: head × (head + 1) × 100 + tail 1 × tail 2.
Example 1, 23×27 =2×(2+1)×103×7=621.
Example 2, 62×68 =6×(6+1)×100 +2×8 =4216.
As can be seen from Example 1 and Example 2, the multiplication of two-digit numbers with the same beginning and the same end can be calculated quickly in this way.
Use the product of "head × (head + 1)" as the first half of the product, and use the product of "tail 1 × tail 2" as the last two digits of the product.
If "Tail 1×Tail 2" is less than two digits, fill in the tens digit with 0.