5,050
Oh, what a lovely math question! If we add 100, 99, 98, 2, 1, and 0.3 together, we get a beautiful total of 300.3. Just imagine each number as a happy little tree in a serene forest of mathematics.
1 + 99 = 100 2 + 98 = 100 3 + 97 = 100 4 + 96 = 100, etc untll you got to 49 + 51 = 100 There would be 49 of these "=100" answers (That would be 4,900) Plus, you would have 50 left over with nothing to match up to. So your total would be 4,950.
There are infinitely many sets: For example, -99, -98, -97, ... -2, -1, 0, 1, 2, ... 97, 98, 99, 100 or 1, 1, 1, 1 ... [100 times] or 0.0001, 0.0001, 0.0001, ... [ a million times].
The number of times that 2 can go into 99 evenly is determined by dividing 99 by 2. This division results in 49 with a remainder of 1. Therefore, 2 can go into 99 a total of 49 times with a remainder of 1.
Consider that 1+99=100 and 2+98=100 you can continue all the way to 49+51=100. This means you have (49x100)+50 (as fifty was left at the end) this gives a final answer of 4950
Oh, what a lovely math question! If we add 100, 99, 98, 2, 1, and 0.3 together, we get a beautiful total of 300.3. Just imagine each number as a happy little tree in a serene forest of mathematics.
303
505
1 + 98 = 99
99 + 1 = 100
1 +100 = 101 100 - 1 = 99 99 +1 = 100 100 times 100 =10000 10000 divide by 2 = 5050Ans:5050
100
99+1=100
1 + 99 = 100 2 + 98 = 100 3 + 97 = 100 4 + 96 = 100, etc untll you got to 49 + 51 = 100 There would be 49 of these "=100" answers (That would be 4,900) Plus, you would have 50 left over with nothing to match up to. So your total would be 4,950.
100
100
printf ("%s\n", "1, 2, 3, 4, 5, 6, ..., 98, 99, 100"); printf ("%s\n", "100, 99, 98, 97, ..., 3, 2, 1");