Okay, let's see. The problem is about Rick and Steve each thinking of a positive factor of 14 and 42 respectively, and they both are thinking of the same number. We need to find how many possible numbers they could be thinking of. Hmm.

First, factors of a number are the numbers that divide it without leaving a remainder. So, for 14, I need to list all its positive factors. Let me start there. 14 is a smaller number, so its factors should be manageable. Let's see: 1, 2, 7, 14. Right? Because 1*14=14, 2*7=14, and those are all the pairs. So factors of 14 are 1, 2, 7, 14.

Now, Steve is thinking of a positive factor of 42. Let me list those as well. 42 is a larger number, but its factors should also be finite. Let's break it down. Starting with 1, 2, 3, 6, 7, 14, 21, 42. Wait, let me check that. 1*42=42, 2*21=42, 3*14=42, 6*7=42. So factors are 1, 2, 3, 6, 7, 14, 21, 42. Yep, that's all. So factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42.

Now, the problem states that both Rick and Steve are thinking of the same number. So we need to find the common factors between the factors of 14 and the factors of 42. In other words, the intersection of the two sets of factors. So let's see:

Factors of 14: {1, 2, 7, 14}
Factors of 42: {1, 2, 3, 6, 7, 14, 21, 42}

The common numbers here would be the numbers that appear in both sets. Let's check each number in the factors of 14:

1: Yes, 1 is in both.
2: Yes, 2 is in both.
7: Yes, 7 is in both.
14: Yes, 14 is in both.

So all four factors of 14 are also factors of 42. Wait, is that true? Let me confirm. 1 divides 42, yes. 2 divides 42, yes. 7 divides 42, yes. 14 divides 42, yes. So all factors of 14 are indeed factors of 42. Therefore, the intersection is all four numbers. So the possible numbers they could be thinking of are 1, 2, 7, and 14. That makes four numbers.

But wait, let me double-check. Maybe I missed something. Let's list the factors again for both numbers:

14: 1, 2, 7, 14
42: 1, 2, 3, 6, 7, 14, 21, 42

Looking at the intersection, numbers present in both lists are 1, 2, 7, 14. So that's four numbers. Therefore, the answer should be 4 possible numbers.

Alternatively, another way to approach this is by finding the greatest common divisor (GCD) of 14 and 42. The GCD would be the largest number that divides both, which is 14. Then, the number of common factors is equal to the number of factors of the GCD. Let's verify that.

Factors of GCD(14,42) = 14. The factors of 14 are 1, 2, 7, 14. So four factors. So that confirms the previous result. Therefore, there are 4 possible numbers they could be thinking of.

Wait, but just to make sure there's no mistake in the factors. Let me check once again. For 14, factors are definitely 1, 2, 7, 14. For 42, factors are 1, 2, 3, 6, 7, 14, 21, 42. Yes, all the factors of 14 are included in the factors of 42. So the intersection is exactly the factors of 14. Therefore, the count is 4. So the answer is 4.

I think that's solid. Both methods—the direct comparison and the GCD approach—lead to the same conclusion. So I'm confident the answer is 4.
Therefore, the number of possible numbers they could be thinking of is $\boxed{4}$.
</think>