**From New Scientist #1278, 5th November 1981** [link]

The fated three sat in “Ye Ink and Blott”

To settle when to carry out the plot.

Said Fawkes “I have this day my passage set:

Before the month is ended I’ll have fled.

Now tell me John and Thomas, if you please,

How many days you’ve free before I leave.”

So John did now the wanted figure state

(The days he’d counted many times of late!)

Tom said he’d three-quarter many days as John

On which he could be there to set the bomb.

“And Tom, I’ve three-quarter many days as thee —

Therefore a handsome choice of dates there’ll be.”

But Guy was wrong since there was just *one* day

On which all three could be there for the fray.

(*One* was the smallest number which there could

Have possibly been with the figures as they stood.)

And hence Guy Fawkes did solemnly decree

“The fifth day of November it shall be.”

Tell me, how many days did there remain

Before the day Guy planned to sail away?

This puzzle is a reference to Guy Fawkes, a conspirator in the Gunpowder Plot, the failure of which is traditionally celebrated in Britain on 5th November.

[enigma134]

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I found this one a little confusingly worded, but I interpreted it as follows.

Guy Fawkes leaves in N days, and is available for “doing the deed” on G of those days. Tom is available on T of those days. John is available on J of those days. And G/T= T/J = ¾. So G < T < J ≤ N. We need to find N given that for any possible availabilities (G, T, J) from N days there must be at least one day where G, T and J are all available.

The analysis doesn't leave much for a program to do, but here's some Python code that runs in 32ms.

Solution:Guy Fawkes leaves in 18 days.