Explanation:

Correct option:3

Solution:

Given that both 421 and 427 leave a remainder of 1 when divided by a certain number, we can express this mathematically as:

421=k*d+1
427=m*d+1
where d is the divisor and k and m are integers. Subtracting these two equations gives:

427−421=(m*d+1)−(k*d+1)

Thus:

6=(m−k)*d

This implies that  d must be a divisor of 6.

Finding the Divisors: The positive divisors of 6 are:1, 2,3,6
Since we are interested in divisors d that satisfy the given condition (i.e., leave a remainder of 1 when dividing both 421 and 427), we can check each divisor: For  d=1: This is trivial, as any number divided by 1 leaves a remainder of 0, so it does not satisfy the condition.
For 
d=2: 
421÷2=210 with a remainder of 1 and 
427÷2=213 with a remainder of 1. Thus, 
d=2 is a valid divisor.
For 
d=3: 
421÷3=140 with a remainder of 1 and 
427÷3=142 with a remainder of 1. Thus, 
d=3 is a valid divisor.
For 
d=6: 
421÷6=70 with a remainder of 1 and 
427÷6=71 with a remainder of 1. Thus, 
d=6 is a valid divisor.
Conclusion: The possible divisors that can be used to get the same remainder 1 are 2, 3, and 6.