Can this be generalized? If the younger paints M digits for every N of
the older (M < N), can a simple formula or algorithm find the number of houses?
Two painters are painting plaques for the numbers of the houses of Puzzle
Street (which are consecutive, starting with 1.)
In the time it takes the younger one to paint 4 digits, the older one
paints 5 digits. The younger begins with the plaques 1, 2, 3, ...
and the older with the last ones.
They paint their last house plaque at the same time and each one has
painted the same number of plaques. How many houses are on the street?
- Same question when the younger paints 5 digits for every 6 of the older.
- Same question when the younger paints 3 digits for every 4 of the older.
- Same question when the younger paints 2 digits for every 3 of the older.
Source: 1. Tangente n° 67 - Gilles Cohen, Alain Fily,Dominique Souder,
submitted by Philippe Fondanaiche. 2-4. Original extensions.
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