Write 1, 2,... On the blackboard As long as there are two or more numbers on the blackboard, erase any two of them a and B, and write A-B (where a ≥ b), and ask whether the last one on the blackboard is odd or even

Write 1, 2,... On the blackboard As long as there are two or more numbers on the blackboard, erase any two of them a and B, and write A-B (where a ≥ b), and ask whether the last one on the blackboard is odd or even

even numbers
In 1,2 In 2003, there are 1001 even numbers and 1002 odd numbers
① Suppose that when any two numbers a and B are even numbers, the resulting A-B (where a ≥ b) is even. When the original even numbers are erased (there is one original even number on the blackboard), 500 even numbers are written, so there are 501 even numbers on the blackboard. Let's erase the even numbers first. In this way, we get 250 new even numbers, and there are 251 even numbers, By analogy, there are even numbers: 126, 63, 32, 16, 8, 4, 2, 1. Now there is only one even number and 1002 odd numbers. At this time, we erase the odd numbers and get even a-b. when 1002 odd numbers are erased, there are 501 even numbers left. In addition, there is a total of 502 even numbers left. Using the same method, we can see that the last one is even
② Suppose that if any two numbers a and B are even, A-B (where a ≥ b) is even. The reasoning method is the same as above. Finally, even numbers are obtained
③ Suppose that the odd and even of any two numbers a and B are erased, and the resulting A-B (where a ≥ b) is odd. When the original number is erased (there is one odd number), 1001 odd numbers are obtained, so that a total of 1002 odd numbers can be obtained. The reasoning is the same as above, and the final even number can be obtained

On the blackboard, there are 1 ~ 2000 zhe 2000 numbers. Each time, perform the following operations: erase two numbers, and write their numbers and,
What is the last one digit left?

The sum of the numbers and the sum of the numbers = the sum of all the numbers of their sum, so there is
1+2+3+..+2000= (1+2000)*2000/2 =2001000
2+0*5+1=3
A: the last digit is 3

There are 1-2013 numbers on the blackboard, two of which can be erased at a time
And write down the sum of these two numbers. If we know that there are four numbers left on the blackboard, and the product is 27, then the sum of these four numbers?

The remainder of a number divided by 9 is equal to the remainder of the sum of the number divided by 9. Each operation changes the sum of the numbers into the sum of the numbers, and does not change the remainder of the number divided by 9. 1 + 2 +. 2013 = 2013x2014 / 2 = 1004x2013 divided by 9 and 3. The sum of the remaining four numbers divided by 9 and 3. 27 = 1x1x1x27 = 1x1x3x9 = 1x3x3, and are 30, 14 and 13 respectively. Only 30 divided by 9 and 3, so the sum of the four numbers is 30