421 and 427, when divided by the same number, leave the same remainder 1. How many numbers can be used as the divisor in order to get the same remainder 1?

The correct option is 3.

To find how many numbers can be used as the divisor such that both 421 and 427 leave a remainder of 1, we can set up the algebraic representation of the division.

Let the divisor be denoted by $d$ (where $d$ is a positive integer) and the remainder be 1. According to the problem, when 421 and 427 are divided by $d$, they both leave a remainder of 1. This means:
421 = $d\times q_1+1$ for some integer quotient $q_1$, and
427 = $d\times q_2+1$ for some integer quotient $q_2$.

Subtracting 1 from both sides of each equation, we get:
420 = $d\times q_1$
426 = $d\times q_2$

This implies that the divisor $d$ must be a common factor of both 420 and 426. Additionally, for a division to yield a remainder of 1, the divisor $d$ must be strictly greater than the remainder. Therefore, we must have:
$d>1$.

Now, let us find the common factors of 420 and 426 by calculating their greatest common divisor (GCD):
The difference between the two numbers is:
426 - 420 = 6.
Since any common divisor of 420 and 426 must also divide their difference, the common divisors must be factors of 6. The factors of 6 are:
1, 2, 3, and 6.

We check which of these factors of 6 can serve as the divisor $d$. Since the divisor must satisfy the condition $d>1$ to leave a remainder of 1, we exclude 1.

The remaining possible values for the divisor $d$ are:
2, 3, and 6.

Let us verify these three divisors:
- If the divisor is 2: 421 = 2 × 210 + 1, and 427 = 2 × 213 + 1 (both leave remainder 1).
- If the divisor is 3: 421 = 3 × 140 + 1, and 427 = 3 × 142 + 1 (both leave remainder 1).
- If the divisor is 6: 421 = 6 × 70 + 1, and 427 = 6 × 71 + 1 (both leave remainder 1).

Thus, there are exactly 3 numbers (2, 3, and 6) that can be used as the divisor to get the same remainder of 1.