Challenge of the Week: The Diya Vendor's Tricky Count

Every festival season, the little lane behind the temple turns into a sea of flickering clay lamps. This week's teaser comes straight from that bustle — a diya vendor with a counting headache and a deadline. He can feel there's a neat trick hiding in his numbers, but he can't quite see it. Can you? Grab a chai, switch off the calculator, and try to crack it with pure reasoning before the timer runs out.

The Challenge

Meena runs a diya stall at the festival market. She has a single tray of clay lamps — fewer than 100 of them — and she keeps trying to arrange them in neat rectangular blocks for display.

She notices something odd:

• When she arranges them in rows of 4, exactly 3 lamps are left over.
• When she arranges them in rows of 5, exactly 4 lamps are left over.
• When she arranges them in rows of 6, exactly 5 lamps are left over.

No matter how she stacks them, she always ends up just one lamp short of a perfect block.

How many diyas does Meena have on her tray?

(There is exactly one possible answer below 100. No tools — just think.)

Hint

Look at the gap between the remainder and the row size each time: 4 − 3 = 1, 5 − 4 = 1, 6 − 5 = 1. In every case she is one short. What happens if you imagine adding a single extra diya to her tray?

Solution

1. Spot the pattern. In each arrangement the remainder is exactly one less than the row size. Rows of 4 leave 3 (one short of 4), rows of 5 leave 4 (one short of 5), rows of 6 leave 5 (one short of 6). So Meena is always one lamp away from a clean block.

2. Add an imaginary diya. Suppose we lend Meena one extra lamp, giving her a tray of (number + 1). Now the leftovers vanish: that new total divides evenly into rows of 4, into rows of 5, and into rows of 6 with nothing left over.

3. Translate into maths. A number that is divisible by 4, 5 and 6 at the same time must be a multiple of their lowest common multiple (LCM).

4. Find the LCM of 4, 5 and 6. Break each into prime factors:

   • 4 = 2 × 2
   • 5 = 5
   • 6 = 2 × 3

   Take the highest power of each prime that appears: 2² (from 4), 3 (from 6) and 5 (from 5). Multiply: 4 × 3 × 5 = 60. So (number + 1) must be a multiple of 60: it could be 60, 120, 180, and so on.

5. Subtract the borrowed lamp. Take back our imaginary diya. The real total is one less than a multiple of 60 — that is, 59, 119, 179, …

6. Apply the limit. Meena has fewer than 100 diyas. The only value below 100 in that list is 59.

7. Check it works.

   • 59 ÷ 4 = 14 rows of 4, remainder 3 ✓
   • 59 ÷ 5 = 11 rows of 5, remainder 4 ✓
   • 59 ÷ 6 = 9 rows of 6, remainder 5 ✓

   All three conditions are satisfied, and 59 + 1 = 60 is exactly the LCM — confirming our "one short" insight.

The charm of this puzzle is the mental flip in step 2. Chasing three separate remainders feels messy and slow. But the moment you notice that every leftover is one short of a full row, the problem collapses into a single clean question: what number becomes perfectly divisible by 4, 5 and 6 the instant you add one? After that, it's just a quick LCM and a subtraction. A lovely reminder that the hardest-looking counts often hide the simplest trick — you just have to look at the gap instead of the pile.

Answer: 59 diyas.