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?

1. **Recall PYQ Pattern:** The question asks for a common divisor leaving the *same* remainder. This reminds me of the 'Method of Differences' used in the 2020 sticks question [3]. The logical shortcut is that the divisor must divide the difference of the two numbers.\n\n2. **Apply Shortcut:** \n Difference = 427 - 421 = 6.\n The potential divisors must be factors of 6.\n\n3. **List Factors:** The factors of 6 are 1, 2, 3, and 6.\n\n4. **Apply Boundary Condition:** As emphasized in the 2023 analysis [29], we must check if these factors are valid. The question states the remainder is 1. Therefore, the Divisor must be strictly greater than 1 ($D > R$).\n\n5. **Elimination:** \n - Factor 1: Invalid (since $1 \\ngtr 1$).\n - Factors 2, 3, 6: Valid (all are > 1).\n\n6. **Final Count:** There are 3 valid numbers (2, 3, and 6).