On the first day of Christmas my true love sent to me a partridge in a pear tree. So how many presents did I get from my true love by the twelfth day of Christmas? This is the sort of problem my family gets tortured with at this time of year by their mathematical father. But it also illustrates some classic traits of the mathematical mind.

There is, of course, the tedious method: counting through the days of Christmas, keeping a tally of all the presents. “OK, on the second day you've got two turtle doves and another pear tree. So that's three presents, plus the first pear tree ....” But soon there will be so many lords a-leaping and species of birds flying round that it will be hard to keep track of everything.

Mathematicians are lazy beasts at heart, so we like a solution that avoids too much computation. Surprisingly, most mathematicians are not that hot on mental arithmetic, preferring a clever bit of lateral thinking to lengthy sums. One classic trick is to limber up by solving a simpler version of the problem. Conveniently, my family also celebrates Chanukkah (although, as Richard Dawkins's successor as the Simonyi Professor for the Public Understanding of Science, I probably should not be admitting that). But this Jewish festival does provide a slimmed-down version of the Christmas problem, and makes a good mathematical warm-up.

On Sunday evening, the first day of Chanukkah, my children will light two candles. The next evening they will light three - and so on until on the eighth day, when they will light all nine candles on the candelabra. So how many candles are needed for the whole festival?

To make things slightly easier, let's add an extra day at the beginning, with one candle. So we want to calculate 1 + 2 + 3 + ... + 9 candles. The slow way is to add up the numbers one by one; the mathematical way is to change the sum into something else. Arrange the candles in a triangle, with one at the top and nine along the bottom. Make a copy of this triangle, but lay it upside-down, with nine candles in the top row and one in the bottom. If we join these triangles together, every row now has ten candles. There are nine rows, making a total of 9 x 10 = 90 candles. But we took two triangles, so we need to halve the number to get 45 candles. Remember that we added an extra day to Chanukkah to make the calculation easier. So we need 44 candles to celebrate Chanukkah.

This trick of turning numbers into geometry is a powerful weapon in attacking such problems, and the 12 days of Christmas poser requires three-dimensional geometry to solve it. As the presents arrive each day, you want to stack them to make a pyramid of presents rather than a triangle. The first layer contains one pear tree; the second a pear tree and two turtle doves; the last a pear tree in one corner and 12 drummers drumming down the long side of the triangle.

If we take six of these pyramids, they can be put together to make a rectangular box 12 presents high, 13 presents long and 14 presents deep. So the total number in each pyramid is 12 x 13 x 14/6 = 364, one present for every day of the year, bar one.

Having solved the two-dimensional Chanukkah problem and three-dimensional Christmas one, the mathematician in me is on the search for another religious festival my family could celebrate that might have a four-dimensional version of our counting problem: the Hindu festival of Diwali, or African festival of Kwanzaa? Or maybe I'll have to start my own fourdimensional cult in hyperspace.

Stuck for a Christmas present? Why not get a symmetrical shape in hyperspace named after a loved one in exchange for a donation to Common Hope, a charity that provides education, housing and healthcare to Guatemalan street kids; visit firstgiving.com/findingmoonshine

THE CONUNDRUM
One hundred guests attended the Mathematical Institute Christmas party. How many handshakes must take place for all 100 guests to shake hands with each other?

Answer: 4,950. This is actually the Chanukkah problem in disguise. The first guest must shake hands with the other 99 guests. The second must shake hands with 98 guests (he has already shaken the first guest's hand). Using the same trick as with the triangle of candles, the total number of handshakes is 99 + 98 + ... +2 + 1 = 99 x 100/2 = 4,950.

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