**From New Scientist #1189, 10th January 1980** [link]

George recently hit upon a plan to get his three daughters up earlier in the mornings. He promised each one penny on each day she was down before him. He further offered one extra penny for a second consecutive day, two extra for a third consecutive day and so on. Thus five consecutive days would be worth 1 + 2 + 3 + 4 + 5 = 15 pence. But the days had to be consecutive.

The scheme ran for eight days, Monday to Monday inclusive. George managed to be first down on one day and third on two others. Otherwise he was always triumphantly last. Alice made more money than Brenda, who was last down on Sunday, and the whole scheme, as it turned out, could not have cost George less than it did for a tally of one first and two third places.

On which day or days was George down before Cynthia?

[enigma46]

### Like this:

Like Loading...

*Related*

In order to arrive at the published solution, I found it necessary to interpret the phrase “the whole scheme could not have cost George less that it did for a tally of one first and two third places” to mean that overall George must have one first place,

nosecond places, two third places andfivelast places.This Python program tries all possible combinations where G is up 1st once, 3rd twice and last the rest of the time in order to find the minimum payout, and also records possible solutions (where A earns more than B, and B is up last on Sunday). Then finally outputs the minimal solutions. It runs in 49ms.

Solution:George was down before Cynthia on Thursday and Saturday.