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I thought I'd explain the logic behind it a bit more, so people are reassured that (extremely) clear thinking is all you need, even if you know little number theory.
The question provides us with the following information:
x = y.k | x ≥ 2, y ≥ 1
The question is: how many values may y assume?
i.e. how many factors does x have?
f.e.
Suppose x = 20
y could be 1 or 20, since 20 = 1. 20
y could be 2 or 10, since 20 = 2. 10
y could be 4 or 5 since 20 = 4. 5
for a total of 6 possible factors.
This does seem like a bit of work though, especially for higher values of x.
Suppose x = 36
How do we determine all distinct factors?
We simply break up 36 into factors and then those factors into factors untill they cant be broken up anymore.
36 = 6 . 6 = (3 . 2) . ( 3 . 2) = 2^2 . 3^2
Both 2 and 3 are primenumbers and primenumbers have the intrinsic property that they cannot be broken up into factors anymore (beside 1 and the prime itself).
So 36 = 2 . 2 . 3 . 3
But from this we know all the factors of 36, because those primes can build any factor of 36.
36 = 2 . 2. 3 . 3 = 2 . (2.2.3) = (2.2) . (3.3) = (2.3).(2.3) etc.
So to know the different factors of a number x, you should break down x into its primefactors. All different possible combinations of those primefactors are the building blocks for all the factors of x.
For 36, the question is:
How many different numbers are 'hidden' in 2.2.3.3?
2.2.3.3 = 2^2 . 3^2
2.2.3 = 2^2 . 3^1
2.2 = 2^2 . 3 ^0
2.3.3 = 2^1 . 3 ^2
2.3 = 2^1 . 3^1
2 = 2^1. 3^0
3.3 = 2^0 . 3^2
3 = 2^0 . 3^1
1 =2^0 . 3^0
So 3*3 = 9 different factors because you can use both primefactors 2 and 3, in three different ways: multiply by 2 or 3 either once, twice or not at all.
3 possible ways to use 2 * 3 possible ways to use 3 = 9 possible ways total.
Note that the actual values of the primefactors are irrelevant in determining how many factors there are in total. You only need to know the power to which those primefactors are raised.
So in statement 1) for a^2 * b^3, there are:
3 possible ways of using 2 (0 times, 1 times, 2 times)
4 possible ways of using 3 (0, 1, 2 and 3 times)
and the total amount of factors is 12.
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Now that's a lot of explaining, but the logical reasoning is actually quite easy to grasp:
*We (should) know primenumbers can build any number
**So if we know the primefactors of x, we can construct all factors of x simply by multiplying those primefactors in all possible ways
To know all possible ways (but not all values), statement 1) suffices
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