Hourglasses, Diophantine equations and Bézout

1. Find the GCD using Euclid

Initially a = 1a + 0b and b = 0a + 1b. For each new remainder subtract q times the divisor row from the dividend row: its coefficients become (u₀ − qu₁, v₀ − qv₁). The last nonzero remainder is the GCD.

2. Check divisibility

3. Bézout and a first solution

Multiply both coefficients by d/GCD:

4. All integer solutions

A negative coefficient means subtracting durations, not negative elapsed time or turning a glass “backwards”. Equation solutions alone are not instructions for operating the glasses.

5. Watch a verified protocol

If a glass of duration a has r minutes remaining, turning it leaves a − r minutes: the sand that already fell returns to the top.

Playback advances one event per second, not one real minute. Sand and remaining times show the state immediately after the move. You can also select table rows.

All events