**Geometry with complex numbers**

What are the coordinates of the reflection of the point (1,0) in the line ?

A Freddo and fame for every well-reasoned attempt at a solution.

Answers to Mr Thomas by Friday 3 November.

**Maths prize 13 October: Pythagoras**

You know 3,4,5 is a “Pythagorean triple”: these are whole numbers which could be the sides of a right-angled triangle. You probably know that 5,12,13 is a Pythagorean triple, too.

You notice that 3×4×5 is even, and 5×12×13 is even, too.

Your tasks:

1. Prove that if a, b, c is a “Pythagorean triple” of whole numbers, then abc must be divisible by 2.

2. In fact there are many numbers bigger than 2 which can be proved to divide exactly into abc for all “Pythagorean triples”. Find some more “Pythagorean triples”, and make a good guess what those numbers are which divide exactly into abc for them all.

3. Prove your guess.

A Freddo (or vegan chocolate) and fame for every good attempt at a solution. Please bring your solution to Mr Thomas by Friday 13 October.

**Maths prize 29 September 2017: factorials**

Jeffrey Sylvester won the prize by getting most of this out.

Full solution at https://mathsmartinthomas.wordpress.com/2017/09/30/maths-prize-for-29-september-2017-factorials/.

*n!* is pronounced “*n* factorial” and means

For example,

• *n!* is the number of possible ways of arranging *n* different items. For example, if there are 30 students in a class and 30 chairs, then there are 30! possible seating plans. Work out 30! on your calculator.

• List these formulas in order of how fast they increase when *n* is big and gets bigger and bigger:

• Triangular numbers (e.g. ) are to addition what *n!* is to multiplication. Work out a quick test you can do on a calculator to check whether a given number *n* is a triangular number without having to calculate all the triangular numbers up to that size. Use your test to find out which of 2278 and 2277 is a triangular number (one of them is).

• 1 = 1! = T_{1} is both a factorial and a triangular number. Using your quick test, find two other numbers which are both factorials and triangular numbers. (They’re not very big). Nobody yet knows whether there are other numbers, apart from those three, which are both factorials and triangular numbers. A check on a computer has shown there are none less than 10^{77337}.

• Primes get thinner on the ground as numbers get bigger, but not very, very thin: it has been proved that for every number *n* > 1, there is always a prime *p* in the range *n < p < 2n*. Use that fact to prove that *n!* can never be a square number if *n*>1

• isn’t a square number, but you can get a square number by removing one term from the multiplication. Which one? Generalise to prove that for any *n* >2 you can get a square number from by removing one term from the multiplication, and say which one.

**Maths prize 15 September 2017: primes**

No entries for this one. I put it down to the effects of the start at the new Sixth Form building.

You check whether 2017 is a prime by checking whether smaller numbers other than 1 divide into it. The shortcuts are:

(1) you need only check for primes, since if say 15 were a factor then 3 and 5 would have been;

(2) you need only check for primes less than √2017, i.e. ≤ 43, since if a prime bigger than 43 is a factor, then it must go fewer than 43 times into 2017, so another prime less than 43 must be a factor;

(3) you can make shortcuts when checking divisibility for particular numbers. Take 43 for example. 2150 is divisible by 43, because 2150=43×50. So 2017 is divisible by 43 only if 2150−2017=133 is divisible by 43, which it obviously isn’t.

The probability of a number chosen at random in the neighbourhood of n (if n is a large number) being prime is about 1/ ln(n).

So, about 1 in 14 for a number around 1082017, and about 1 in 8 for a number around 2017.

I worked those out in my head by saying, for example, that log_{10}2017 is about 3.3, and log_{10}e is about 0.43, so ln 2017 is about 3.3/0.43, which is about 8.

How can you give a quick answer as to whether 8099 is prime?

You can think of the difference of two squares: 8099 = 8100−1 = (90−1)(90+1) = 91×89, so not prime.

Or you could think: if there is a quick answer at all, it must be that 8099 is not prime. There is no quick way (quicker than the way I used for 2017 in the hospital) to prove that it is prime, but there can always be a quick way to prove that a number is not prime – just find a factor.

**Maths prize 21 July 2017**

**Skyscrapers**

The New York World Building (above) was New York’s first skyscraper, in 1890. Lots more were built in the next few decades.

Why weren’t skyscrapers built before? It wasn’t because builders couldn’t build tall: Old St Paul’s Cathedral, in London, was 149 metres tall, much taller than the World Building (106 metres). It wasn’t because lifts couldn’t be installed. Lifts had been included in buildings since the 1850s. According to John J Carty, first head of the research wing of the Bell Telephone Company, it was the invention and spread of the telephone that made skyscrapers a going proposition. Why?

Genie Louis de Canonville won the prize for the best attempt at a mathematical explanation, with some help from the other Year 12 Further Maths students. Click here for report.

**Maths prize 29 June 2017**

**A train on a circular track**

A train goes round a circular track at a varying speed, but in a constant pattern (i.e. its speed at each point on the track is the same every time it goes round). Prove there must be at least one pair of diametrically-opposite points on the track at both of which the train has exactly the same speed.

The prize was won by Mohaned al-Bassam

Fortnightly maths prize puzzles were suspended between February and the AS and A2 exams in May-June.

**Maths prize 10 February 2017**

Jetmir Guri and Mohaned al-Bassam won prizes for these puzzles about the number 23, designed to mark my younger daughter Molly’s 23rd birthday.

1. When Molly was little, her mum had a rule that Molly could have a birthday party only every even year. Suppose instead the rule was that Molly could have a party only every prime-number birthday. At what age would she first have to wait two years for her next birthday party? At what age does she first have to wait four years? At what age will she first have to wait six years?

2. At which age would she first have a party that was more than two years away from any other party (i.e. more than two years from the previous party, and more than two years from the next)?

3. Suppose Molly decided (or, when she was little, her mum decided) that each year she would celebrate her birthday on a date chosen at random. It’s unlikely her 2nd birthday would be the same date as her 1st, but if she lives to 366 it’s certain that the same date will have been the birthday-celebration day twice. (Ignore leap years). At what age does the probability of the same date having been the birthday-celebration date twice rise above 50%?

4. What is the smallest number which cannot be written as the sum of 8 or fewer cubes? (1=1^{3}, 2=1^{3}+1^{3}, 17=2^{3}+2^{3}+1^{3}, 15=2^{3}+1^{3}+1^{3}+1^{3}+1^{3}+1^{3}+1^{3}+1^{3}, etc.)

**Maths prize 27 January 2017**

You choose one card out of 21 without telling the magician which one.

The 21 cards are dealt out face up in three columns of 7 cards each. You tell the magician which column contains your card.

The magician picks up the cards, column by column, so that the column with your card is in the middle.

She or he deals them out across three columns again, top card in first column, 2nd card in 2nd column, 3rd card in 3rd column, 4th card in 1st column, etc. Then the same process (tell which column, pick up, deal again). Then you tell which column a third time. The magician picks up and identifies your card as the 11th card.

Explain mathematically how this is done. See if you can generalise to different numbers of cards. Bolaji Atanda and Mohaned al-Bassam won prizes: see report.

**Maths prize 13 January 2017**

How many square numbers have no even digits?

Bolaji Atanda won the prize: click here for report.

**Maths prize 16 December 2016**

Find the *exact* total

Can you find a similar sum that adds up to 5?

Can you find a similar sum with a total which is not a whole number?

Prizes were won by Bolaji Atanda and Genie Louis de Canonville. Click here for report.

Maths prize for 25 November 2016: Goat in a field

A goat is tethered to the outside of a square barn in the middle of a large field. Where on the barn should the tether be attached to give the goat the biggest grazing area? Prizes were won by Conner Lake and Bolaji Atanda. Mention should also go to Rashi Bhatt and Genie Louis de Canonville, who worked on it and made some progress but in the end didn’t submit an entry.

Maths prize for 11 November 2016: Kinder eggs

In one series of Kinder Surprise chocolate eggs, sold at airports, each egg contains a toy Airbus plane, in one of five liveries. If the toys are distributed randomly among the eggs, what is the expected number of eggs you must buy to get a complete set? Bolaji Atanda and Mohaned Al-Bassam won prizes for completely correct solutions. Click here for report.

**Maths prize for 21 October 2016: Benford’s Law**

Students were asked to investigate Benford’s Law. Maz Razwan won the prize. Click here for report.

**Maths prize for 7 October 2016: Arithmetic progressions and primes**

What is the pattern of the differences of trios of primes bigger than 3 which are in arithmetic progression?

All the trios have differences divisible by 6. The difference may be 6, as in 11, 17, 23, or 24, as in 23, 47, 71, but it is always divisible by 6.

Alex On won the prize. Click here for report.

**Maths prize for 23 September 2016: Remainders and clock arithmetic**

Bolaji Atanda won the prize. Click here for report.

**Maths prize for 15 July 2016: Treasure Island**

Captain Bluebottle has buried his treasure on Triangle Island, and the secret message says only that the treasure is in the exact centre of the island. Where is that?

There are a number of possible alternative answers.

Vinh Chu, David Trieu, Wei-kong Mao, and Alex On won prizes. Click here for a report.

**Maths prize for 1 July 2016: Crossing the desert**

You want to cross a huge desert. You have an unlimited supply of Toyota pick-up trucks, drivers, and fuel. Each pickup can travel only 360km on a full tank, and the desert is too dangerous to leave fuel dumps or for pickups or drivers to wait or stay in the desert. How wide a desert can you cross?

Alex On worked out the answer but didn’t write it up, so no prize this time. Click here for the solution.

**Maths prize for 24 March 2016**

Two questions:

**The first question: is 3599 a prime?**

**Second question: a spider is at the centre of a base of a box which is a 1-metre cube. A fly is at one of the top corners. The spider can’t fly, but only walk on the surfaces inside the box. What is the shortest distance by which the spider can reach the fly?**

How do you answer them *quickly*? Hamse Adam, Mugisha Uwiragiye, and Dion Miller found full solutions, and Taija Williams a partial answer. Click here for solution.

**Maths prize for 11 March 2016: fast cube roots**

Ask a friend to think of a two-digit number, cube it, and tell you the answer. Then you can tell them the cube root straight off, without having to do any calculation.

How? Look at the last digit of the number they’ve given you. If they’ve given you 250047, for example, that’s 7.

The second digit of the cube root is the same, except that 2 swaps with 8 and 3 swaps with 7. If the last digit is 2 then the second digit of the root is 8; if the last digit is 3 then second digit of the root is 7; and vice versa.

So for 250047, last digit is 7, second digit of root is 3.

Then forget about the last three digits of the number you’re given, and just look at the first ones: 250~~047~~ gives you 250 in this case.

The first digit of the root is the biggest digit with a cube smaller than what you’ve now got. In this case, 6, because 6^{3}=216, but 7^{3}=343, which is bigger than 250.

In this example, cube root of 250047 is 63.

Your challenge:

1. Why isn’t there a similar easy way to find square roots?

2. What other roots is there a similar easy method for (with two-digit numbers)? 4th roots? 5th roots? 6th…?

Joan Onokhua and Hamse Adam found solutions, and the Year 12 Further Maths class also found the solution collectively, with Jetmir Guri providing most of the insights. Click here for report.

**Maths prize for 12 February 2016**

You have a lot of identical books, stacked on the edge of a table. How far from the table-edge can the pile reach without falling over? This is an important issue in construction, for example in building arches above entrances.

No-one submitted a prize-winning entry. Alex On and Khaleah Edwards deserve honourable mention for getting most of the way there, but neither wrote it up.

**Maths prizes for 2 February 2016**

There were two prize puzzles for 2 February.

Conner Lake and Alex On solved the “22” puzzle:

• Archimedes discovered 22/7 as a good fractional approximation to π over 2200 years ago, without the help of algebra or calculators. To what % is the approximation accurate?

• A semiprime is a number which is made by multiplying together exactly two prime numbers; a semiprime pair is two numbers next to each other which are both semiprimes, like 14 and 15. What is the second semiprime pair? (No-one knows yet whether there are infinitely many semiprime pairs: the best result so far is that there are infinitely many couples of semiprimes differing by 26 or less).

• How many pieces can you cut a pizza into with just six cuts?

No-one claimed the double prize, to prove that 22/7 overestimates π by proving

Lola Behanzin, Taija Williams, and Serene Williams solved the “Morley Miracle” puzzle: to prove that in any triangle, if you draw the lines dividing each angle into three equal parts (the trisectors), they meet in the middle to create an equilateral triangle.

Click here for Lola’s, Taija’s, and Serene’s proof.

**Maths prize for 19 January 2016**

You have 45 carrots, and want to get a donkey to take as many as possible of them to a endpoint 15 miles away. Trouble is, the donkey can only carry 15 carrots at a time, and insists on eating a carrot early in every mile. How many carrots can you get to the endpoint?

The prize was won by Wei-Kong Mao, with a correct answer of 8 carrots. Click here for details.

**Maths prize for 15 December 2015**

You have 12 billiard (or snooker) balls, all the same except one which is slightly heavier – or it may be slightly lighter – than the rest.

You have scales which you can use to compare balls (or groups of balls) with each other.

How can you identify the odd ball by using the scales only three times?

Wei-Kong Mao, Mai Nguyen, and Zion Mills won prizes. Click here for report.

Maths prize for 1 December 2015

The painting above, done in 1895, is of Russian peasant boys doing a mental arithmetic puzzle. As you can see, one of them has the answer and is whispering it into the teacher’s ear.

The puzzle, shown on the blackboard, is:

(10^{2} +11^{2} +12^{2} +13^{2} +14^{2}) / 365

Like the Russian boys, you have no calculator and no paper. Prize for:

1. A good reasoned guess at the answer

2. The exact answer, with an explanation of how you got it by mental arithmetic.

Vinh Chu, Tariq Hall, Sharif Quansah, Dion Miller, and Mugisha Uwiragiye solved this, and won prizes. Click here for a range of solutions

**Maths prize for 17 November**

An old ideas about checking addition sums of whole numbers is “casting out nines”. Go through each number to be added, and add its digits, ignoring any nines. If your answer is more than nine, subtract nine, and keep on going until you have a number less than nine.

Add up all the less-than-nine numbers you get that way.

Calculate the similar less-than-nine number for your answer to the original addition sum. It should be the same.

The example it gives is 59+38=97. The less-than-nine numbers for the two numbers being added are 5 and 2. 5+2=7 = less-than-nine number for 97. It checks out.

Your challenge:

1. Explain why this works.

2. Would it work if we took, say, 7, instead of 9? If it would, why cast out nines rather than sevens?

The prize was won by Alex On: click here for solution.

**Maths prize for 3 November 2015: weighing**

You have a scale with two pans, so for example you can measure 2 grams exactly by putting a one-gram weight in one pan and a three-gram weight in the other. With what four weights can you measure any weight up to 40 grams? With what five can you measure any weight up to 121 grams?

The prize (200 Vivos and a Freddo) was won by Khaleah Edwards. Click here for solution.

**Maths prize for 13 October 2015**

If you have a group of four people, it’s possible to have no subgroup of three all of whom know each other, and simultaneously no subgroup of three none of whom know each other. **If there are five people at a party, is it possible that the party includes no subgroup of three who all already know each other, and no subgroup of three all of whom didn’t know each other before the party? If there are six, is that possible?**.

The prize was won by Joan Onokhua: click here for more.

**Maths prize for 29 September 2015: profit or loss?**

The task was to make a well-explained assessment of the claim by Hutchison, the world’s biggest container-terminal operators, that they were making losses at their terminal in Brisbane; and that they were making more losses the more containers they moved.

It was the first-ever maths prize to have no entries, other than 23 June 2015 when students were worn out after exams. Some fortnights previously, no-one got a prize because no-one wrote down their solution, or half-solution, but this is the first time no student has had any solution at all.

The lesson, evidently, is we should stick with the simply-stated problems requiring mathematical ingenuity, and not bring in the more real-life-type problems which involve wading through masses of information to extract the key figures needed for a good calculation.

Click here for the information given for the problem, and an answer.

**The maths prize question for 15 September 2015 was a Conway puzzle: Assemble six 2x2x1 blocks and three 1x1x1 blocks into a 3x3x3 cube.**

Eight students solved this problem, and could explain how by word of mouth: Hamse, Khaleah, Jetmir, Mugisha, Lou-Lou, Matthew, Lola, and Sharif. Mugisha went one better and wrote out the maths of how to solve the problem, so he gets an extra prize.

The maths prize question for 18 July was: How far apart should the centres of the two circles be for the three areas A, B, and C to be equal?

Mugisha Uwiragiye, Sharif Quansah, Dion Miller, and Juan Florez solved this together, with Mugisha making the decisive breakthrough. Click here for report.

The maths prize for 7 July was won by Mugisha Uwiragiye. Problem: A friend comes to visit. She or he tells you the number of the bus they’ve come by. You say: “Aha. That’s the smallest number which can be written as the sum of *squares* in two different ways”. What was the number of the bus? What was the number of the bus if it was the smallest number which can be written as the sum of *non-zero* squares? Answers: 25 and 50. Click here for report.

The maths prize question for 23 June was:

**Part 1:**

To help you: 1+2+3+4+….+n=½n(n+1) for all n

**Part 2:**

If for n = 3 the pairwise sum is 1×2 + 1×3 + 2×3, which makes 11, find a general formula which works for any n.

No-one produced an answer to this one, which I guess is due to it being post-exam time rather than it being specially difficult. Click here for answers.

24 March was the last fortnightly prize before the exams. We start again in the second half of June.

For the 24 March prize (arithmetic with remainders), Sharif Quansah and Hayden Leroux won 200 Vivos and a Freddo.

No winners for the prize for 24 February: If all the people of the world stood shoulder-to-shoulder (with the babies in slings, and the old and sick people held up by the people next to them), how much area would we cover? The year 13 further maths students *told* me an answer, but none of them *handed it in*. The answer is: about the area of a large city: click here for the calculation.

See here for report on the 3 February prize – to prove that the crescent-moon shaped region has the same area as the shaded triangle. Mariama Bah, Shannon Bradley, and Sharif Quansah won prizes.

See here for report on the 20 January prize question (Pythagorean triples). Hamse Adam won the prize.

See here for report on 16 December prize question (Turing’s last sigh). No-one claimed the prize, though Soner Hasan got close to working it out.

See here for report on the 9 December prize question (the shape-sorter with the circular shape fitting exactly into the right-angled triangle). No-one won a prize: Daniel Huang got the solution, but didn’t write it up.

See here for report on 25 November prize question (the seating plans). Sharif Quansah, Mugisha Uwiragye, and Hamse Adam all won prizes.

See here for report on 11 November prize question (the cored apple). The Year 12 further maths class won the prize with a collective answer.

See here for report on 26 October prize question (whether to tie your shoelaces on static ground or on a moving walkway). Deniz Yukselir won the prize.

See here for 10 October prize question (regions in a circle). Hamse Adam, Deniz Yukselir, and Hayden Leroux all won prizes.

See here for 26 September prize question (which numbers = the sum of two square numbers). Deniz Yukselir had worked out the answer to the first bit (which odd primes are the sum of two square numbers), and Daniel Huang had 90% worked out that answer, but neither wrote up his answer in time.