Shortcut Trick
The problem states that the required least number x must be perfectly divisible by 7.
We can efficiently find the correct answer by checking the given options for divisibility by 7.
Checking options: 3506 ÷ 7 (No) | 3560 ÷ 7 (No) | 3065 ÷ 7 (No).
Only 3605 ÷ 7 = 515, which yields a remainder of 0.
∴ The correct answer is 3605.
Alternate Method
Given:
Divisors = 12, 15, 18, 20 and 25
Remainder in each case = 5
The least number x is perfectly divisible by 7.
Formula Used:
Required Number = (LCM of divisors) × k + Remainder
Calculations:
⇒ First, we find the LCM of the given divisors: 12, 15, 18, 20 and 25.
⇒ 12 = 22 × 3
⇒ 15 = 3 × 5
⇒ 18 = 2 × 32
⇒ 20 = 22 × 5
⇒ 25 = 52
⇒ LCM = 22 × 32 × 52 = 4 × 9 × 25 = 900
⇒ Let the required number be x = 900k + 5, where k is a positive integer.
⇒ Since x is divisible by 7, the expression (900k + 5) must be a multiple of 7.
⇒ We can split 900k as (896k + 4k). Here, 896 is a multiple of 7, since 896 = 7 × 128.
⇒ Therefore, the divisibility depends entirely on (4k + 5) ÷ 7 leaving a remainder of 0.
⇒ Let's check for minimum integer values of k:
⇒ For k = 1: 4(1) + 5 = 9 (Not divisible by 7)
⇒ For k = 2: 4(2) + 5 = 13 (Not divisible by 7)
⇒ For k = 3: 4(3) + 5 = 17 (Not divisible by 7)
⇒ For k = 4: 4(4) + 5 = 21 (Perfectly divisible by 7)
⇒ Substitute k = 4 into the original equation x = 900k + 5.
⇒ x = 900 × 4 + 5 = 3600 + 5 = 3605.
∴ The correct answer is 3605.
Additional Information
LCM and Constant Remainders
When a number is divided by a, b, and c leaving the same remainder R in each case, the general form of the required number is represented as: N = k × LCM(a, b, c) + R.
LCM and Different Remainders
When a number is divided by a, b, and c leaving remainders x, y, and z respectively, such that the difference (a − x) = (b − y) = (c − z) = K, the general form of the number is: N = k × LCM(a, b, c) − K.
Divisibility Rule of 7
To check if a number is divisible by 7, drop the last digit, double it, and subtract it from the remaining truncated number. If the result is exactly divisible by 7 (or is 0), then the original number is also divisible by 7.