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?

• If 421 and 427 leave a remainder of 1 when divided by a number $d$, then $d$ must divide $(421 - 1)$ and $(427 - 1)$.

• This simplifies to finding the common divisors of 420 and 426.

• The difference between the two numbers is $427 - 421 = 6$, so $d$ must be a divisor of 6.

• The divisors of 6 are 1, 2, 3, and 6.

• We must check which of these divisors actually produce a remainder of 1 for both numbers.

• Divisor 1: Remainder is always 0, so it is invalid.

• Divisor 2: $421 \div 2$ gives remainder 1; $427 \div 2$ gives remainder 1. Valid.

• Divisor 3: $421 \div 3$ gives remainder 1 ($4+2+1=7$, $7 \div 3$ rem 1); $427 \div 3$ gives remainder 1 ($4+2+7=13$, $13 \div 3$ rem 1). Valid.

• Divisor 6: $421 \div 6$ gives remainder 1 ($420$ is divisible by 6); $427 \div 6$ gives remainder 1 ($426$ is divisible by 6). Valid.

• Thus, the valid divisors are 2, 3, and 6, making a total of 3 numbers.