Let s be the largest number that divides 1005, 2025, and 6235, leaving the…

2025

Let s be the largest number that divides 1005, 2025, and 6235, leaving the same remainder in each case. What is the sum of the digits of s?

  1. A.

    1

  2. B.

    7

  3. C.

    8

  4. D.

    9

Show answer & explanation

Correct answer: A

When several numbers leave the SAME remainder on division by some number s, then s must divide every pairwise difference of those numbers exactly - subtracting any two of them cancels out the common remainder, leaving a multiple of s. So the greatest possible s is the HCF (highest common factor) of all the pairwise differences.

  1. Find the pairwise differences: 2025 minus 1005 = 1020; 6235 minus 2025 = 4210; 6235 minus 1005 = 5230.

  2. Express each difference as a product of primes: 1020 = 22 × 3 × 5 × 17; 4210 = 2 × 5 × 421; 5230 = 2 × 5 × 523.

  3. The common prime factors across all three differences are 2 and 5, so HCF = 2 times 5 = 10. Hence s = 10.

  4. Sum of the digits of s: 1 + 0 = 1.

Cross-check: dividing each original number by 10 leaves the same remainder in every case (1005, 2025 and 6235 all leave remainder 5 on division by 10), confirming s = 10 is valid. Since 10 is exactly the HCF of the differences, no larger number can satisfy the condition, so it is the greatest possible value.

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