Question 89

Mathematics Binomial Theorem Hard

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

(A) 123
(B) 127
(C) 235
(D) 305
✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Mathematical Principle

The problem asks for the greatest number 'n' that divides 1657 leaving a remainder of 6, and divides 2037 leaving a remainder of 5. This is a problem that can be solved by finding the Highest Common Factor (HCF).

The underlying principle is:

  • If a number 'a' divided by 'n' leaves a remainder 'r1', it means that (a - r1) is perfectly divisible by 'n'.
  • If a number 'b' divided by 'n' leaves a remainder 'r2', it means that (b - r2) is perfectly divisible by 'n'.
To find the greatest such number 'n', we need to find the HCF of (a - r1) and (b - r2).

Step 1: Subtract the Remainders

First, we subtract the respective remainders from each number to find the numbers that our required divisor 'n' must divide perfectly.

  • 1657 - 6 = 1651
  • 2037 - 5 = 2032

Now, we need to find the HCF of 1651 and 2032.

Step 2: Find the HCF using the Euclidean Algorithm

We can use the division method to find the HCF:

  1. Divide the larger number by the smaller number:
    2032 = 1 × 1651 + 381
  2. Divide the previous divisor (1651) by the remainder (381):
    1651 = 4 × 381 + 127
  3. Divide the previous divisor (381) by the new remainder (127):
    381 = 3 × 127 + 0

The last non-zero remainder is the HCF. In this case, the HCF is 127.

Conclusion

The greatest number which satisfies the given conditions is 127.

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