V

Vaibhav Shukla • 4.53K Points
Extraordinary Math

  • (A) 123, 20
  • (B) 113, 10
  • (C) 116, 20
  • (D) 118, 15
Correct Answer - Option(B)

Explanation by: Vaibhav Shukla
The greatest common divisor = HCF of (3739 – 2270), (6677 – 3739) and (6677 – 2270)
= HCF of 1469, 2938 and 4407
Now, 1469 = 1469 × 1
2938 = 1469 × 2
4407 = 1469 × 3
∴ HCF = 1469
Now, 1469 = 113 × 13
Since, (2270 – R), (3739 – R) and (6677 – R), where R is the remainder, are exactly divisible by 1469, hence these are also exactly divisible by its factors 13 and 113. The three digit number is 113. Now the above mentioned numbers can be written as
2270 = (113 × 20 ) + 10
3739 = (113 × 33) + 10
6677 = (113 × 59) + 10
Hence, the required number is 113 and the remainder is 10.

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