Shortcut Trick
The required number must be of the form [LCM(8, 16, 18, 20, 25) × k] + 3.
LCM of the given divisors is 3600, so the number x = 3600k + 3.
Since x is exactly divisible by 7, (3600k + 3) ÷ 7 must yield a remainder of 0.
3600 ÷ 7 leaves a remainder of 2, so we only need to check the remainder of (2k + 3).
For k = 2, 2(2) + 3 = 7, which is fully divisible by 7. Thus, x = 3600 × 2 + 3 = 7203.
∴ The correct answer is 7203.
Alternate Method
Given:
Divisors = 8, 16, 18, 20 and 25
Remainder in each case = 3
The number x is exactly divisible by 7.
Formula Used:
Required Number = LCM(Divisors) × k + Remainder
Calculation:
⇒ Prime factorization of the given divisors:
⇒ 8 = 23
⇒ 16 = 24
⇒ 18 = 2 × 32
⇒ 20 = 22 × 5
⇒ 25 = 52
⇒ LCM = 24 × 32 × 52
⇒ LCM = 16 × 9 × 25 = 3600
⇒ Therefore, the required least number x is of the form 3600k + 3.
⇒ We are given that x is exactly divisible by 7.
⇒ 3600k + 3 = (514 × 7 + 2)k + 3
⇒ 3600k + 3 = 7 × 514k + (2k + 3)
⇒ For the entire expression to be completely divisible by 7, the remaining term (2k + 3) must be a multiple of 7.
⇒ Put k = 1: 2(1) + 3 = 5 (Not divisible by 7)
⇒ Put k = 2: 2(2) + 3 = 7 (Exactly divisible by 7)
⇒ Substitute k = 2 into the general expression for x:
⇒ x = 3600 × 2 + 3
⇒ x = 7200 + 3 = 7203
∴ The correct answer is 7203.
Additional Information
LCM Remainder Concept (Same Remainder)
When a number divided by a, b, and c leaves the same remainder r in each case, the required least number is given by LCM(a, b, c) × k + r.
LCM Remainder Concept (Different Remainders)
When a number divided by a, b, and c leaves remainders p, q, and r respectively such that (a − p) = (b − q) = (c − r) = d, the required number is LCM(a, b, c) × k − d.
HCF Remainder Concept (Same Remainder)
The greatest number that divides a, b, and c to leave the same remainder in each case is the HCF of |a − b|, |b − c|, and |c − a|.
Co-prime Numbers
Two numbers are said to be co-prime if their HCF is exactly 1. The LCM of two co-prime numbers is always equal to their product.