Archive for Mathematics

Xmas Maths Challenge!

Hello, Mysci! It has been a long time, but I’m back! Hooray! Anyhoo…
My new school asked me to complete an Xmas maths challenge, so here it is (minus footnotes and a table):

Xmas Maths Challenge

Questions

Every Christmas, Santa delivers presents to every good child in the world. If this were the case:

1) How many houses would he visit?

2) How far would he travel…assuming he’d like to keep it to a minimum?

3) How long could he spend at each house?

4) How fast would he travel if he only delivers between dusk and dawn?

5) If he has a piece of christmas cake and a glass of sherry at each home, how
much does he eat and drink altogether?

6) If he were breathalysed, how many times over-the-limit would he be?

7) Rudolph eats a carrot at each house. How much would he weigh at dawn?

Ext) If he delivers a single present to every child in a box measuring 50cm
x 30cm x 20cm, how much wrapping paper and sellotape would he need?

Answers

1) There are 1.9 billion kids in the world, but some of them are bad. If we assume that 1 billion of the 1.9 billion kids are good, and about 300 million of them live with another good kid, then Santa should visit about 700 million houses. These would be all over the world. Santa would probably need clones to cope with such a large amount of houses. This is a nail in the coffin for the Santa Claus theory. This text is basically going to totally disprove the existence of Santa.

2) This problem is related to the travelling salesman problem, which is VERY hard to solve. To maximise time, Santa should definitely start (and finish!) on the international date line. This would give him 36 hours at most, because if he starts at dusk (6pm) then by the time he has gone around the world, following the sun, it will be 6pm again. Now, he has another 12 hours, in total making 36 hours. Of course, since Santa lives at the North Pole, he should also start and finish there. However, if we assume the Australian view, that Santa lives at the South Pole, He should start and finish there instead. Santa can also skip the major unpopulated areas, such as Antarctica, the Sahara Desert and the Amazon Rainforest. Now if we assume it’s 2,200km to the nearest house on the IDL from whichever pole Santa starts at, and it is only 1m between houses in the same town and Santa visits all 3,000,000 towns in the world, and it’s at most 5km between each, and there are 233 houses per town, then Santa would travel 714,004,400km. This is, naturally, utterly ridiculous.

3) Assuming Santa set aside 18 of his 36 hours for visiting houses, that would be 64,800 seconds for houses. Divide this by 700 million and you get 0.0000925 seconds per house. Even if he set aside 24 hours for houses, he’d only be able to spend 0.00012 seconds per house. He would barely have time to drop a present in, let alone eat or drink any cake or sherry. He’d have to do it en route or something else similar. The theory that Santa goes down the chimney can, therefore, be disproved.

4) Assuming Santa set aside 18 hours for travel, or 64,800 seconds (I will keep throwing that data around!), then he’d have to travel 11,018.58 km/s or 11,018,580 m/s to make the entire 714,004,400km. For reference, the speed of light is 299,792,458 m/s and Santa’s speed is just over 1/27 of that. Humans still haven’t invented technology to go that fast! It’s even worse if Santa sets aside only 12 hours (43,200 seconds) he’d have to travel at 16527.87km/s or 16,527,870 m/s, or 1/18 of the speed of light. This, of course, is utterly ridiculous, ridiculous enough, in fact, that Special Relativity applies. However, for simplicity, I have left out that factor.

5) The average volume of a glass of an alcoholic beverage is 180ml, but they are usually only filled to 100ml. Assuming that each glass is filled only to the usual amount, and there was nothing but sherry offered, Santa would drink 70,000,000,000L of sherry. A bottle of sherry is 750ml, and it costs £10. Santa would drink 93,333,333,333 bottles of sherry and the public would spend £933,333,333,330. I am NOT paying for drinks. Following the same method, a piece of christmas cake’s mass is 60g, plus 10g for the icing and decoration. If every christmas cake in the world were the same, and Santa ate all 700 million slices, he’d have consumed 4,900,000kg of christmas cake in all. If each cake cost £10, and a cake had eight slices, then the public would end up spending £612,5000 on christmas cake. In total, Santa would eat 357,000,000kg of cake and drink 70,000,000,000L of sherry, and the public would spend £933,333,945,8330 on Santa. And don’t get me started on the cost of Rudolph’s carrots!

6) It is, unfortunately, illegal (and impossible) to breathalyse somebody who is dead, as Santa is likely to be, unless his metabolism is extremely quick-acting. Assuming the latter, Santa would only be around a million times over the limit, which is 0.08% (according to my source, Google), as opposed to around one million one thousand times assuming the former, which would mean certain death. All in all, it’s best to stick to the old milk-and-cookies approach, to be on the safe, not drunk side.

7) To solve this apparently simple question, I measured 10 carrots, and listed their weights (on earth!) on Table 1.The average of their weights was 1.041 Newtons. Multiply this by the number of houses (700 million), and then add the average weight of a reindeer. According to my sources (centrally Wikipedia), and assuming Rudolph is a Finnish forest reindeer (R. t. Fennicus), the average weight of a male reindeer is 1961 Newtons. If Rudolph is a female reindeer however, her weight (before the carrots) would be 980 Newtons. Assuming Rudolph is male, then after the carrots he’d weigh 728,701,961 Newtons, but if she’s a girl, then she’d be 728,700,980 Newtons. Add the extra weight for his/her glowing nose (0.05 Newtons), and he/she’d be 728,701,961.005 Newtons (or 728,700,980.005 Newtons if she’s a female) in total. However, in space (where no-one can hear you scream) he/she would weigh barely anything. Because you asked for weight, I have listed all weights in Newtons, the only proper weight measurement. Some of these values are simply ridiculous. Another nail in the coffin of the Santa Claus theory, then.

Ext) A piece of sellotape is 5 centimetres long. To use the minimum amount of wrapping paper, Santa would need 12 pieces per box, meaning that 60 centimetres of sellotape would be needed per box. In total, that’s 600,000,000 metres of sellotape. A roll of Scotch® tape is 33m long, so Santa would use 18,181,819 rolls of tape if he used Scotch®. If a roll costs him 75p, he would end up spending £13,636,364.30 for tape alone. Because the surface area of the box is 630 centimetres squared, to use the minimum amount of packaging, Santa would need two 50cm x 30cm panels, two 50cm x 20cm panels and two 30cm x 20cm panels. If every child gets one gift, then he would use 630,000,000,000cm squared of wrapping paper or 6,300,000,000m squared.

Therefore, THE ANSWER is… is… 42.