Question

An equal number of laddoos have been placed in 3 different boxes. The laddoos in the first box were distributed among 20 children equally, the laddoos in the second box among 24 children, and those in the third box among 12 children. Not a single laddoo was leftover. Then, what was the minimum number of laddoos in the three boxes altogether?

Answer 1

The lowest common multiple of 20, 24 and 12 gives the minimum number of laddoos in one box.

Multiples of 20 = 20, 40, 60, 80, 100, 120, 140, 160, 180, 200

Multiples of 24 = 24, 48, 72, 96, 120

Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120

∴ LCM of 20, 24 and 12 = 120

∴ Minimum number of laddoos in 1 boxes = 120

∴ Minimum number of laddoos in 3 boxes = 3 × 120 = 360

∴ The minimum number of laddoos in 3 boxes are 360.

Answer 2

Step 1: Prime factorization of the numbers

• 20 = 2² × 5
• 24 = 2³ × 3
• 12 = 2² × 3

Step 2: Identify the highest powers of all prime factors

• For 2, the highest power is 2³.
• For 3, the highest power is 3¹.
• For 5, the highest power is 5¹.

Step 3: Calculate the LCM

LCM = 2³ × 3¹ × 5¹

= 8 × 3 × 5 = 120

Thus, each box contains 120 laddoos, since it is the minimum number divisible by 20, 24, and 12.

Step 4: Total laddoos in three boxes

Total laddoos = 120 × 3

= 360