18. 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 ?

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Accepted Answer

The correct answer is option C. If two numbers, N1 and N2, leave the same remainder 'r' when divided by a divisor 'd', then (N1 - r) and (N2 - r) must be exactly divisible by 'd'. Here, N1 = 421, N2 = 427, and r = 1. So, (421 - 1) = 420 and (427 - 1) = 426 must be divisible by the common divisor 'd'. This means 'd' must be a common factor of 420 and 426. First, find the prime factorization of each number: 420 = 2^2 × 3 × 5 × 7. 426 = 2 × 3 × 71. The common factors are found by taking the minimum powers of the common prime factors. The common prime factors are 2 and 3. The HCF (420, 426) = 2^1 × 3^1 = 6. The common divisors of 420 and 426 are the factors of their HCF, which are 1, 2, 3, and 6. However, a divisor must always be greater than the remainder. Since the remainder is 1, the possible divisors must be greater than 1. Therefore, the numbers that can be used as the divisor are 2, 3, and 6. There are 3 such numbers. This question tests number theory, specifically properties of remainders and HCF.

Q18. 18. 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 ?

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