Why 105?
A pile of 105 = 3 × 5 × 7 coins fills every row exactly. Adding or removing 105 coins changes no leftover, so the answers repeat every 105.
33 / ANTIQUITY TO THE RENAISSANCE
Qin Jiushao
Count a pile of coins in 3s, 5s and 7s and note what is left over each time. Those leftovers are enough to find the pile, once you have one helper number for each divisor.
Explore the mathematics ↓
01 / SEEKING UNITY
One number must fit every clue.
Change the pile until all three leftovers match the clues.
Then seek unity: find a helper that leaves 1 for one divisor and nothing for the others, and build any answer from the helpers.
Clues:
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THE IDEAFor 3, 5 and 7 the helpers are 70, 21 and 15. Take each as many times as its leftover, add, and remove 105s: 2 × 70 + 3 × 21 + 2 × 15 − 2 × 105 = 23.
The person behind the ideas
- Published Mathematical Treatise in Nine Sections in 1247.
- Worked on numerical equations and surveying.
- Wrote zero as a small circle in his treatise.
More of the mathematics
Shared factors
Clues can contradict each other when divisors share a factor: a count that leaves 3 in fours is odd, but one that leaves 4 in sixes is even. Qin’s procedure deals with such divisors before assembling a solution.