The answer process is as follows:

1-2=- 1; 3-4=- 1; 5-6=- 1 until 99- 100=- 1, because there are 100 numbers, every two numbers are a group, so a * * * has 50 groups of differences of-1.

How much is that? 1-2+3-4+5-6+...+99- 100

=( 1-2)+(3-4)+(5-6)+...+(99- 100)

=- 1 x 50

=-50 ?

Extended data:

Algorithm of integer addition and subtraction:

(1) Same-digit alignment;

(2) from the unit;

(3) When it is added to dozens, it will be pushed to a higher position; If the subtraction is not enough, subtract 1 from the high position. When this number is added to 10, it will decrease.

Addition attribute:

From additive commutative law's associative law, it can be concluded that when several addends are added, the positions of addends can be exchanged at will; Or add a few addends first and then add them with other addends, and the sum remains the same. For example: 34+72+66+28 = (34+66)+(72+28) = 200.

Break ten methods

For example, if we calculate 13-5, the first step is to divide 13 into 10 and 3. We know that 10-5 equals 5, and then 5 plus 3 finally equals 8. So13-5 =10+3-5 =10-5+3 = 5+3 = 8.