To solve this problem, we need to use the concept of modular arithmetic. Modular arithmetic deals with the remainder when a number is divided by another number, called the modulus.

Let's assume the two numbers are x and y, and the divisor is d.

The given conditions can be represented as:
x ≡ 3 (mod d) -- (1)
y ≡ 4 (mod d) -- (2)
(x + y) ≡ 2 (mod d) -- (3)

To find the value of the divisor, we need to consider the possible values of d.

1. Divisor as 3:
If d = 3, then the possible remainders when divided by 3 are 0, 1, and 2. However, the remainders given in the problem are 3 and 4, which are not possible when divided by 3. Therefore, we can rule out the divisor as 3.

2. Divisor as 7:
If d = 7, then the possible remainders when divided by 7 are 0, 1, 2, 3, 4, 5, and 6. We can check if the given remainders satisfy the conditions.

From (1), x ≡ 3 (mod 7) implies x = 7a + 3 for some integer a.
From (2), y ≡ 4 (mod 7) implies y = 7b + 4 for some integer b.
Substituting these values in (3), we get:
(7a + 3 + 7b + 4) ≡ 2 (mod 7)
(7a + 7b + 7) ≡ 2 (mod 7)
7(a + b + 1) ≡ 2 (mod 7)
7(a + b + 1) = 7k + 2 for some integer k.

Since the left-hand side is divisible by 7, the right-hand side should also be divisible by 7, which is not possible for 7k + 2. Therefore, we can rule out the divisor as 7.

3. Divisor as 5:
If d = 5, then the possible remainders when divided by 5 are 0, 1, 2, 3, and 4. We can check if the given remainders satisfy the conditions.

From (1), x ≡ 3 (mod 5) implies x = 5a + 3 for some integer a.
From (2), y ≡ 4 (mod 5) implies y = 5b + 4 for some integer b.
Substituting these values in (3), we get:
(5a + 3 + 5b + 4) ≡ 2 (mod 5)
(5a + 5b + 7) ≡ 2 (mod 5)
5(a + b + 1) ≡ 2 (mod 5)
5(a + b + 1) = 5k + 2 for some integer k.

Since the left-hand side is divisible by 5, the right-hand side should also be divisible by 5. 5k + 2 satisfies this condition for certain values of k. Therefore, the divisor can be 5.

Hence,