Shortcut Trick
The smallest number which leaves a remainder r when divided by a, b, and c is given by: LCM(a, b, c) + r
⇒ Required number = LCM(9, 12, 15) + 3
⇒ Required number = 180 + 3 = 183
∴ The correct answer is 183.
Alternate Method
Given:
Divisors = 9, 12, and 15
Common remainder in each case = 3
Formula Used:
Required Number = LCM(divisors) × k + Remainder
For the smallest possible number, we take k = 1.
Calculations:
Find the prime factorization of each divisor to calculate the LCM:
⇒ 9 = 32
⇒ 12 = 22 × 3
⇒ 15 = 3 × 5
The Least Common Multiple (LCM) is the product of the highest powers of all prime factors involved:
⇒ LCM(9, 12, 15) = 22 × 32 × 5
⇒ LCM(9, 12, 15) = 4 × 9 × 5
⇒ LCM(9, 12, 15) = 180
Now, add the common remainder to obtain the required smallest number:
⇒ Required number = LCM(9, 12, 15) + 3
⇒ Required number = 180 + 3 = 183
∴ The correct answer is 183.
Additional Information
General Form of Such Numbers
Any number that leaves a remainder r when divided by divisors a, b, and c is written in the form: (LCM(a, b, c) × k) + r, where k ≥ 1 is an integer.
LCM with Different Remainders (Constant Difference)
If a number leaves remainders p, q, and r when divided by a, b, and c respectively, such that (a − p) = (b − q) = (c − r) = k (constant difference), then the smallest such number is: LCM(a, b, c) − k.
HCF with Remainders
To find the greatest number that divides x, y, and z leaving remainders p, q, and r respectively, calculate: HCF(x − p, y − q, z − r).