Producing a list of factor pairs is the easiest way.
1 x 100
2 x 50
4 x 25
5 x 20
10 x 10
As 10 is duplicated then there are 9 different numbers that are factors of 100.
If if it just a total number of factors that is required (rather than the values) then this can be obtained by stating the original number in terms of its prime factors.
100 = 22 x 52.
Add 1 to each exponent and obtain the product : (2 + 1)(2 + 1) = 3 x 3 = 9 ......which confirms the answer given above.
The limit is infinity if the factors do not have to be whole numbers. If you stipulate that the factors have to be whole numbers, then, yes, for each number, there is a limit to how many factors it has. For example, the number 4 has only 3 whole-number factors: 1, 2, and 4.
1,2,3,6,9,18,27,54
75's factors are: 1,3,5,15,25,75 so it has 6 factors
Three.
14 (I think).
All of them. Different numbers have different numbers of factors.
Four of them.
Four: 1 5 25 125.
Its factors are: 1, 2, 4 and 8 which makes four of them
Six: 1 2 3 4 6 12.
12 different numbers.
Six of them.
The limit is infinity if the factors do not have to be whole numbers. If you stipulate that the factors have to be whole numbers, then, yes, for each number, there is a limit to how many factors it has. For example, the number 4 has only 3 whole-number factors: 1, 2, and 4.
11,1,3,33
1,2,3,6,9,18,27,54
Six numbers. They are 1, 2, 4, 5, 10 and 20.
There are infinitely many such numbers.