# The Indisputable Santa Mathematical Advent CalendarDay 20

Happy 20th of December!

Throughout the month, to accompany the release of our book on the Mathematics of ChristmasHannah Fry & I are tweeting out Christmathsy bits and pieces, one a day, advent calendar style. Assuming we don’t run out of ideas, that is…

OK, since the upcoming final four puzzles of the calendar may be a little tough, today, some light relief…

Answers via the comments or on Twitter. All correct answers will be rewarded with deep respect and warm Christmas wishes. Enjoy!

CLICK HERE TO SEE THE WHOLE CALENDAR SO FAR

SOLUTION TO YESTERDAY’S PUZZLE

Scroll down for the solution…

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Solution:

The answer is 70 turns clockwise.

The number of teeth turned through needs to be equal to the lowest common multiple of the numbers on all the cogs, which is 1260. However, if (like me, I should confess) you initially thought that this was the answer, then you were working the elves work far too hard! One turn of the handle is equal to 18 teeth it itself, so moving through 1260 teeth requires only 1260/18 = 70 turns.

Here’s an alternative explanation from Olaf Doschke (click on the comment to see his suggested video!):

Other correct solutions (for one or both parts of the problem) came in from Jonathan Footprints Bicyle, Mr. Jasinski, Matt Triggs, Jamie is Fed Up, Paul Hardman, Troy ToTheWorld, Martin Benfer, wpswart, Pierre Schramm and Catherine Evans.

There’s a present there for each of you. Though only after you’ve whittled yourselves down to two through a no-holds-barred free-for-all of competitive carolling, since there were only two gifts on the belt.