Shortcut Trick
LCM of 10, 20, and 25 is 100, so the total sweets must be a multiple of 100.
We check multiples of 100 for the condition: (Number ÷ 27) leaving a remainder of 3.
100 ÷ 27 → Remainder is 19.
200 ÷ 27 → Remainder is 11.
300 ÷ 27 → Remainder is 3 (since 300 = 27 × 11 + 3).
∴ The correct answer is 300 sweets.
Alternate Method
Given:
The number of sweets is completely divisible by 10, 20, and 25.
When divided among 27 children, 3 sweets are left.
Concept Used:
Number = LCM(a, b, c) × k
Dividend = Divisor × Quotient + Remainder
Calculations:
⇒ LCM of 10, 20, and 25 = 100
⇒ Let the required minimum number of sweets be 100k, where k is a positive integer (1, 2, 3...).
⇒ According to the second condition, 100k ÷ 27 leaves a remainder of 3.
⇒ This can be expressed as: 100k = 27q + 3
⇒ Since 100 = 27 × 3 + 19, we can simplify the expression by checking only the remainder part: 19k ÷ 27.
⇒ For k = 1: 19(1) ÷ 27 → Remainder is 19
⇒ For k = 2: 19(2) = 38 → 38 ÷ 27 → Remainder is 11
⇒ For k = 3: 19(3) = 57 → 57 ÷ 27 → Remainder is 3 (since 57 = 27 × 2 + 3)
⇒ Thus, the smallest integer value for k is 3.
⇒ Minimum number of sweets = 100 × 3 = 300
∴ The correct answer is 300 sweets.
Additional Information
LCM and HCF Relationship
For any two numbers A and B, the product of the numbers is equal to the product of their LCM and HCF: A × B = LCM × HCF.
Remainder Theorem
Dividend = (Divisor × Quotient) + Remainder. This is the core equation used for solving divisibility and remainder problems.
Common Remainder Rule
The smallest number which when divided by x, y, and z leaves a common remainder r in each case is represented as: LCM(x, y, z) + r.