Shortcut Trick
Since the number x is perfectly divisible by 7, we can directly check the given options for divisibility by 7.
390 ÷ 7 → Remainder 5
930 ÷ 7 → Remainder 6
903 ÷ 7 → Remainder 0 (Exactly divisible)
309 ÷ 7 → Remainder 1
∴ The correct answer is 903.
Alternate Method
Given:
Divisors = 12, 15, 18, 20 and 25
Remainder in each case = 3
The number is exactly divisible by 7.
Concept Used:
The least number which when divided by given numbers leaves a common remainder r is of the form: (LCM of numbers) × k + r, where k is a positive integer (1, 2, 3, ...).
Calculations:
⇒ First, find the LCM of 12, 15, 18, 20, 25 using prime factorization:
⇒ 12 = 22 × 3
⇒ 15 = 3 × 5
⇒ 18 = 2 × 32
⇒ 20 = 22 × 5
⇒ 25 = 52
⇒ LCM = 22 × 32 × 52 = 4 × 9 × 25 = 900
⇒ The required number x must be of the form: 900k + 3
⇒ Given that x is divisible by 7, so (900k + 3) must be a multiple of 7.
⇒ We can rewrite 900k + 3 as: (128 × 7 + 4)k + 3
⇒ 900k + 3 = 128 × 7k + (4k + 3)
⇒ Since 128 × 7k is completely divisible by 7, the remainder term (4k + 3) must also be divisible by 7.
⇒ Let's check values for k:
⇒ If k = 1: 4(1) + 3 = 7 (Which is perfectly divisible by 7)
⇒ So, the least value of x is obtained when k = 1.
⇒ x = 900(1) + 3 = 903
∴ The correct answer is 903.
Additional Information
Same Remainder Concept
The least number which leaves the same remainder 'r' when divided by a, b, and c is given by the formula: LCM(a, b, c) + r.
Different Remainder Concept
If the respective differences between divisors and remainders are the same (i.e., a−x = b−y = c−z = k), the required number is: LCM(a, b, c) − k.
Divisibility Rule of 7
Double the last digit of the number and subtract it from the remaining part. If the result is 0 or a multiple of 7, the original number is divisible by 7.