Boxing Laddoos and Suppliers Who Arrive Together
The owner of a sweet shop in Muzaffarpur has prepared three separate batches for a festival order: 240 besan laddoos, 360 motichoor laddoos and 540 khoya laddoos. He wants to pack each batch in identical boxes, keeping the three varieties unmixed, using the largest possible number of laddoos per box, with no box left partly filled. Separately, three suppliers - milk, khoya and sugar - visit the shop at fixed intervals of 15, 20 and 25 days respectively. All three happened to visit today, and the owner must hire an extra helper on every day when all three come together. The owner tells his accountant that this will happen three times during the next 300 days, and he budgets helper wages on that basis. He also wants a small trial pack in which laddoos can be arranged exactly in rows of 12, 18 or 24, whichever row size he chooses, without any leftover.
1. Find the largest number of laddoos that can be placed in one box, and the total number of boxes needed for the three batches.
Read the worked answer
The box size must divide each of 240, 360 and 540 exactly, and must be as large as possible - so it is the HCF.
Prime factorisation:
- 240 = 2⁴ × 3 × 5
- 360 = 2³ × 3² × 5
- 540 = 2² × 3³ × 5
Take the lowest power of each common prime: 2² × 3 × 5 = 60.
So each box holds 60 laddoos. Boxes: 240 ÷ 60 = 4, 360 ÷ 60 = 6, 540 ÷ 60 = 9. Total boxes = 4 + 6 + 9 = 19.
Check: 60 divides all three counts exactly (remainder 0), and no larger number does, since 120 fails on 540 and 90 fails on 240.
2. State when all three suppliers will next visit on the same day, and decide whether the owner's claim of three such days in the next 300 days is correct.
Read the worked answer
Common visit days occur at multiples of the LCM.
15 = 3 × 5, 20 = 2² × 5, 25 = 5². LCM = 2² × 3 × 5² = 300.
So the next combined visit is on day 300 from today; after that, day 600, and so on.
Counting the next 300 days as days 1 to 300, only day 300 is a combined visit - exactly one occasion, not three. The claim is false, and the helper budget is roughly three times what is needed.
A boundary note worth stating: if today itself is counted as day 0 of a 300-day window, there would be two combined visits (today and day 300). Either way, three is wrong; the answer depends on whether the window is open or closed at today, so the counting convention must be stated.
3. Find the smallest number of laddoos that can be arranged exactly in rows of 12, 18 or 24 with no leftover.
Read the worked answer
The number must be a multiple of 12, of 18 and of 24, and we need the smallest such number - the LCM.
12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3. LCM = 2³ × 3² = 72.
Verification: 72 ÷ 12 = 6, 72 ÷ 18 = 4, 72 ÷ 24 = 3 - all exact.
Since 72 is a common multiple and no smaller number is divisible by 24 and 18 simultaneously (multiples of 24 below 72 are 24, 48; neither is divisible by 18), 72 is the minimum. A trial pack of 72 laddoos works.