There are an infinite number of possible answers.
Think of any number: 2.1358?
calculate: 408 - 2.1358 = 405.8642
and then 2.1358 + 405.8642 = 408
You can do that for any of infinite numbers that you can think of.
Similarly, apart from 0, you can do that with multiples of any number.
The numbers that are divisible by 408 are infinite. The first four are: 408, 816, 1224, 1632 . . .
Sum = ½ × number of numbers × (first number + last number) = ½ × (462 - 408 + 1) × (408 + 462) = ½ × 55 × 870 = 23,925
408 = 2 × 2 × 2 × 3 × 17
Four hundred eight thousand.
Prime numbers are not divisible by any numbers other than themselves and 1.
The numbers that are divisible by 408 are infinite. The first four are: 408, 816, 1224, 1632 . . .
Sum = ½ × number of numbers × (first number + last number) = ½ × (462 - 408 + 1) × (408 + 462) = ½ × 55 × 870 = 23,925
408 = 2 × 2 × 2 × 3 × 17
408
408000000
The range is 408. (777 - 369 = 408)
2 x 2 x 2 x 3 x 17 = 408
Four hundred eight thousand.
408
Prime numbers are not divisible by any numbers other than themselves and 1.
I am asssuming that the numbers are too large for you to simply see them and "know" their GCF. The easiest way to find the GCF of 2 numbers is the Euclidean method. It is somewhat awkward to explain but once understood, is very easy to use. The idea here is to make the numbers that you are dealing with smaller and smaller so as to simplify the problem. Suppose you start with the two numbers p and q where p > q. Assume that they are not equal for if p = q, then their GCF is p (or q). The GCF of p and q is the same as the GCF of the smaller number, (say q), and p-q. Repeat this process and keep going until the two numbers are the same. So, for example, ket us try to find the GCF of 1836 and 1428 GCF(1836, 1428)] = GCF(1428, 1836-1428) = GCF(1428, 408) = GCF(408, 1428-408) = GCF(408, 1020) = GCF(408, 1020-408) = GCF(408, 612) = GCF(408, 612-408) = GCF(408, 204) = GCF(204, 408-204) = GCF(204, 204) The two numbers are the same so STOP! The answer is 204. It is not a particularly fast method, but it is simple: all you need to now is subtraction.
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