Shortcut Trick
To find the required number, simply find the LCM of the given divisors and add the decreased value.
LCM of 15, 25, and 55 = 825
Required Number = 825 + 3 = 828
∴ The correct answer is 828.
Alternate Method
Given:
The numbers are 15, 25, and 55.
The required number when decreased by 3 is exactly divisible by these numbers.
Formula Used:
Required Number = LCM(a, b, c) + k (where k is the value by which the number is decreased)
Calculations:
⇒ Prime factorization of 15 = 3 × 5
⇒ Prime factorization of 25 = 5 × 5 = 52
⇒ Prime factorization of 55 = 5 × 11
⇒ LCM (15, 25, 55) = 3 × 52 × 11
⇒ LCM = 3 × 25 × 11
⇒ LCM = 825
⇒ The smallest number exactly divisible by 15, 25, and 55 is 825.
⇒ Since the required number was decreased by 3, we add 3 back to the LCM.
⇒ Required Number = 825 + 3 = 828
∴ The correct answer is 828.
Additional Information
LCM (Least Common Multiple)
The smallest positive integer that is perfectly divisible by each of the given numbers.
Case 1: Same Remainder
The smallest number which when divided by a, b, and c leaves the same remainder 'r' in each case is given by: Required Number = LCM(a, b, c) + r.
Case 2: Different Remainders
The smallest number which when divided by a, b, and c leaves remainders x, y, and z respectively, such that (a − x) = (b − y) = (c − z) = k, is given by: Required Number = LCM(a, b, c) − k.
LCM of Fractions
LCM of fractions = LCM of Numerators ÷ HCF of Denominators.