Understanding the Problem
We're given that $x \times y=20$, where $x$ and $y$ are positive integers. We need to find out how many different possible values $x+y$ can take based on these pairs.

Finding Factor Pairs of 20
First, let's list all pairs of positive integers $(x, y)$ such that $x \times y=20$. To do this, we'll find all the factors of 20 and pair them up.

The positive factors of 20 are: $1,2,4,5,10,20$.
Now, let's pair them up:
1. $x=1$, then $y=20$ (since $\(1 \times 20=20\)$ )
2. $x=2$, then $y=10($ since $\(2 \times 10=20)\)$
3. $x=4$, then $y=5$ (since $\(4 \times 5=20\)$ )
4. $x=5$, then $y=4$ (since $\(5 \times 4=20\)$ )
5. $x=10$, then $y=2$ (since $\(10 \times 2=20\)$ )
6. $x=20$, then $y=1$ (since $\(20 \times 1=20\)$ )

However, notice that some pairs are just reverses of each other, like $(2,10)$ and $(10,2)$. Since addition is commutative ( $x+y=y+x$ ), these will give the same sum. So, we can consider unique pairs where $x \leq y$ to avoid duplicates in sums:
1. $(1,20)$
2. $(2,10)$
3. $(4,5)$

The other pairs $(5,4),(10,2)$, and $(20,1)$ will give the same sums as the above, respectively.

Calculating the Sums
Now, let's calculate $x+y$ for each unique pair:
1. $1+20=21$
2. $2+10=12$
3. $4+5=9$

So, the possible sums are 21,12 , and 9 .
Checking for Other Possibilities
Wait a minute, are there any other factor pairs we might have missed? Let me list all the factor pairs again to be sure:
- $\(1 \times 20\)$
- $\(2 \times 10\)$
- $\(4 \times 5\)$
- $\(5 \times 4\)$
- $\(10 \times 2\)$
- $\(20 \times 1\)$

No, that covers all combinations where both numbers are positive integers. As mentioned, the sums from the reversed pairs are the same:
- $5+4=9$ (same as $4+5$ )
- $10+2=12$ (same as $2+10$ )
- $20+1=21$ (same as $1+20$ )

So, the distinct sums are indeed 9,12 , and 21 .

Counting the Different Sums
Now, let's list the different sums we've obtained:
1. 9
2. 12
3. 21

That's 3 different possible values for $x+y$.
Verifying the Options
Looking back at the options:
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

We've found there are 3 different possible sums, so the correct answer is $\mathbf{B}$.

Potential Missteps
Initially, one might think to list all ordered pairs ( $x, y$ ), leading to six pairs, but since addition is commutative, $(x, y)$ and $(y, x)$ yield the same sum. Thus, considering unique unordered pairs gives us three distinct sums.

Another possible oversight is missing some factor pairs, but listing them systematically as done above ensures we've captured all possibilities.

Final Answer
After carefully listing all factor pairs and calculating their sums, we find there are $\mathbf{3}$ different possible values of $x+y$.

Answer: B. 3