There are 2003 numbers on the blackboard. Erase two numbers at any time and write another one______ After that, there is only one number left on the blackboard

There are 2003 numbers on the blackboard. Erase two numbers at any time and write another one______ After that, there is only one number left on the blackboard

Wipe two at any time, write another one, reduce the number of 1, do not write the last time, so, need 2003-2 + 1-2 + 1-2 + 1 =2003-1-1 = (2003-2) / (2-1) + 1 = 2002 (Times). A: after 2002 times, there is only one number left on the blackboard

You can double the number written on the blackboard or erase the last number each time
Each time, you can double the number written on the blackboard or erase its last number. Assuming that the number written at the beginning is 458, how can you get 14 through the above changes

458 double 916
916 to mantissa 91
91 times 182
182 double 364
364 to mantissa 36
36 times 72
72 times 144
144 to mantissa 14

The blackboard says 1, 2, 3, 4 498, a total of 498 numbers, each time arbitrarily erase two of them, and write their difference, after several times, there is only one number 0 left on the blackboard, is this possible? Why?

Since the parity of the sum of two numbers and the difference between two numbers is the same, the parity of the difference between two numbers after each operation is the same as that of the original two numbers, so the parity of the difference between the remaining numbers after several times is the same as that of the sum of the original 498 numbers. The sum of the original 498 numbers is (1 + 498) × 498 △ 2 = 499 × 249, which is odd, and 0 is not odd, so there is no such possibility