Shortcut Trick
Required number = HCF of (1356 − 12), (1868 − 12) and (2764 − 12).
Required number = HCF of 1344, 1856 and 2752.
The difference between the first two numbers (1856 − 1344) is 512.
The required HCF must be a factor of 512. From the given options (74, 64, 54, 44), only 64 is a perfect factor of 512.
∴ The correct answer is 64.
Alternate Method
Given:
Numbers given are 1356, 1868 and 2764.
The remainder in each case is 12.
Formula Used:
Required greatest number = HCF of (a − r), (b − r) and (c − r)
Calculations:
⇒ First number = 1356 − 12 = 1344
⇒ Second number = 1868 − 12 = 1856
⇒ Third number = 2764 − 12 = 2752
⇒ Prime factorization of 1344 = 26 × 3 × 7
⇒ Prime factorization of 1856 = 26 × 29
⇒ Prime factorization of 2752 = 26 × 43
⇒ HCF of (1344, 1856, 2752) = 26 = 64
∴ The correct answer is 64.
Additional Information
Case 1: Same Remainder Given
The greatest number that divides a, b, and c leaving a remainder r in each case is the HCF of (a − r), (b − r), and (c − r).
Case 2: Same Remainder Not Given
The greatest number that divides a, b, and c leaving the same remainder in each case is the HCF of |a − b|, |b − c|, and |c − a|.
Case 3: LCM Remainder Concept
The least number which when divided by a, b, and c leaves a remainder r in each case is [LCM of (a, b, c)] + r.