No, they're not. The range from 100 to 1,000 includes numbers like 100.6, 239.07,
5431/2, and 942.638, and those are not 'whole numbers'.
But the numbers in that range that have nothing after the decimal point, and have
no fraction after them, are all whole numbers.
The same is also true of any number that's notbetween 100 and 1,000 .
101
All numbers from 1 to 100 which are whole numbers are integers
5050
The total of all of the numbers from 1 to 99 is 4950.
If you can count from 1 to 100, then you can name them all. (And in numerical order, I might add.)
101
1000 numbers. From 19,500 to 20,499 all round to 20,000
Whole numbers between 1 and 100 are just counting numbers. Can you count? Just start at 1 and count to 100. All the numbers in your count will what you are asking for: That is: 1, 2,3, ... (fill in the missing numbers ) 98,99,100.
All numbers from 1 to 100 which are whole numbers are integers
5050
The total of all of the numbers from 1 to 99 is 4950.
No. Factorials can only be placed on positive whole numbers or 0. For example, 0!, 1!, and 6! are all fine, but 2.5! and (-1)! are not. Since factorials can only be placed on whole numbers, that means you will only be multiplying whole numbers. 100! = 100 * 99 * 98 ... * 1; all those numbers are whole numbers, meaning your answer will be a whole number.
If you count 0000 to 9999 you have 10,000. For whole numbers you have 9999. * * * * * Not sure about "whole" numbers. There are 10,000 in all, of which 9000 (from 1000 to 9999) do not have leading 0s.
I wrote out all the factors of all the numbers from 1 to 100 and counted them.
First, let's find the sum of 100 terms, where a1 is 1 and an = 100. Sn = (n/2)(a1 + an) S100 = (100/2)(1 + 100) = 50(101) = 5,050 The average of all whole numbers from 1 to 100 is: 5,050/100 = 50.5
1100
If you can count from 1 to 100, then you can name them all. (And in numerical order, I might add.)