Q. 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 ?

Explanation :

421 and 427 

Let’s approach this step-by-step:

1) First, we need to understand what the question is asking. We’re looking for a number that, when dividing both 421 and 427, leaves a remainder of 1 in both cases.

2) Mathematically, we can express this as:
421 = k * d + 1
427 = m * d + 1
Where d is our divisor, and k and m are some integers.

3) Subtracting these equations:
427 – 421 = (m – k) * d
6 = (m – k) * d

4) This means that d must be a factor of 6.

5) The factors of 6 are: 1, 2, 3, and 6.

6) Let’s check each:

– If d = 1:
421 ÷ 1 = 421 remainder 0
427 ÷ 1 = 427 remainder 0
This doesn’t work.

– If d = 2:
421 ÷ 2 = 210 remainder 1
427 ÷ 2 = 213 remainder 1
This works!

– If d = 3:
421 ÷ 3 = 140 remainder 1
427 ÷ 3 = 142 remainder 1
This works!

– If d = 6:
421 ÷ 6 = 70 remainder 1
427 ÷ 6 = 71 remainder 1
This works!

7) Therefore, there are 3 numbers (2, 3, and 6) that can be used as the divisor to get the same remainder 1 for both 421 and 427.

The correct answer is c) 3.