According to Lisa Pollack in the Financial Times of 23 December, presumably drawing on some neuro-science research: Continue reading

# Monthly Archives: December 2014

# M2: Work-energy principle and statics

# FP3: matrices and vectors

# C4 maths: vectors

LESSON 1: DRAWING VECTORS

Starter: draw some examples of vectors on the whiteboard, and describe them. Continue reading

# Some quiz problems

These quiz problems are not in any special order, other than being roughly sorted into number-theory, algebra, measure, probability, topology, and geometry problems. Some are very easy, some harder. Almost all are cribbed from somewhere else.

N1: Without a calculator – what is 77×87−76×87?

N2: Without a calculator – what is 101^{2}−99^{2}?

N3: Why can you tell whether a number is divisible by 9 by seeing whether the sum of its digits is divisible by 9?

N4: The tens digit of a square number is odd. What is its units digit?

N5: Two brothers sell their sheep. The price in roubles of each sheep = the number of sheep. Then they divide up the proceeds by each taking 10 roubles in turn. The younger brother ends up short because, the last time it’s his turn to get 10 roubles, there are less than 10 left. His older brother gives him his penknife, and that balances the share exactly. How much is the penknife worth?

N5. During the French Revolution, as well as introducing the metric system for lengths and weights, the revolutionary authorities also decreed that a circle would now be 400 degrees instead of 360. The metric system for lengths, weights, etc. has spread from France to the whole world, but the 400 degree circle did not stick. Why not?

N6. The revolutionary authorities also changed the week from seven days to ten. Why didn’t that stick?

N7. Why 24 hours in a day?

N8. You do Eratosthenes’ sieve by crossing out all the numbers divisible by 2, divisible by 3, divisible by 5, and divisible by 7, and those that remain are all the prime numbers up to 100. Why do you need to check only those four divisibilities?

N9. If you did Eratosthenes’ sieve to find all the prime numbers up to 300, which numbers would you have to cross out all the multiples of?

N10. Ten students of different heights are placed in a row. They could be in order from shortest to tallest, or in order from tallest to shortest. On the other hand, how mixed-up can the order be? What is the smallest possible number of students within the ten which are in the “right” order, i.e. either shortest first, next-shortest somewhere to the right of that, next-but-one shortest somewhere to the right again, etc., or conversely?

N11: Work out 1^{5}, 2^{5}, 3^{5}, 4^{5}, and their remainders when divided by 5. What pattern do you see?

Do you see the same pattern when you calculate the remainders of 1^{7}, 2^{7}, 3^{7}, 4^{7}, 5^{7}, 6^{7}, when divided by 7?

Do you see the same pattern when you calculate the remainders of 1^{9}, 2^{9}, 3^{9}, 4^{9}, 5^{9}, 6^{9}, 7^{9}, 8^{9}, when divided by 9?

Make a good guess about a general rule for the remainder of m^{n} when divided by n (if m is a whole number less than n). The rule is true for a certain category of whole numbers n, but not all.

What rule follows, if n is in that category of whole numbers, for the remainder of m^{n−1}when divided by n?

Assuming your guess is right, show that:

2^{54321}+3^{15432} is divisible by 11.

M1: What is the biggest possible area of a rectangle of perimeter 4?

M2: What is the biggest possible area of a triangle of perimeter 6?

M3. How can you prove that an equilateral triangle has a bigger area than any other triangle of the same perimeter?

M4. Which of mass, time, length, volume, and area can we not measure? (We can only calculate or count it).

M5. Why?

M6. Your floor is 3.03 metres×2.97 metres×2.4 metres×3.6 metres. If carpet tiles are 0.5m×0.5m, and sold in packs of ten, how many packs do you need to cover it?

M7. To calculate the area of the floor more exactly, we’d need to know the angles at the corners. But that’s hard to measure without an “angle gauge protractor” with two long metal arms, costing about £15 or £20. What other measurement, easier and cheaper to take, would enable us to calculate the area exactly?

M8. You’re building a house and want the walls to be at right angles. What cheap method can you use to measure the right angles? (Hint: You do have a rope).

M9. Why can a cat survive a big fall but an elephant can’t?

P1. Name a possible event of probability zero.

P2. A tennis tournament has 2n competitors. How many different ways are there to pair them for the first round?

P3. In a game show a winner gets to choose from three boxes, one of which contains a good prize and the other two a goat (and you don’t want a goat). You choose. The game show host opens one of the boxes you’ve not chosen to show it contains a goat, and gives you a chance to swap choices. Should you swap?

P4. Draw six dots. If each pair of dots is connected by one line, how many lines?

P5. Everyone in France always shakes hands with a friend when meeting them for the first time in a particular day. You and five friends meet for the first time one day. How many handshakes?

P6. An shop offers six flavours of ice cream. How many options for a double-scoop ice cream with two different flavours?

P7. The original Baskin Robbins shops offered 31 flavours. How many double-scoop different-flavours options?

P8. Cold Rock ice cream shops in Australia advertise over 3000 flavour combinations, single-scoop, double-scoop, or triple-scoop. How many basic flavours?

P9. There are 23 students in a class. What’s the probability that two have the same birthday? (Ignore leap years).

P10. There are 30 students in a class, and 30 places in the classroom. How many possible seating plans?

P11. You’re arranging a party, and ten people promise each to take responsibility for one part of the organisation. If any one person fails or delays too much, then the party will have to be cancelled. All ten people are 90% reliable, i.e. there is probability 0.9 that they will do their job reasonably on-time. What’s the risk of having to cancel the party?

P12. Prove every way of shuffling a deck of cards can be reduced to a series of swaps which just swap one card and another. Prove that, although there are many different ways of reducing a shuffle to a series of swaps, if one series for a shuffle has an even number of swaps, then every other series also has an even number. (Which implies if one series for a shuffle has an odd number of swaps, then every other series also has an odd number).

P13. Take a pen. Draw parallel lines on the floor (or on a big sheet of paper, on the floor) which are the same distance apart as the length of the pen. Drop the pen at random on the floor. What is the probability that the pen will cross or touch a line?

A1. You want to get from one end of an airport terminal to the other, but you must tie up your shoelace on the way. Is it quicker to tie your shoelace when you are on a moving walkway or when you are on stationary ground?

A2. Roughly what is the sum from r=1 to r=100 of r^{9}?

A3. Roughly what is 100! ?

A4. Prove that e is irrational.

C1. If f(x) is defined by f(0)=0 and f(x)=x^{2}.sin (1/x) for f(x)≠0, what is f'(0)?

T1. B and C and D can be written without taking your pen off the page or retracing a line, but A can’t. Which other letters can? What is the rule?

T2. Königsberg used to be one of the main cities of Germany. Now it is part of Russia and called Kaliningrad. However, in the old days it had seven bridges connecting the opposite banks of the river Pregel and the city’s two islands. Was it possible to walk through the city crossing each bridge exactly once?

G1. A circle has constant width, but a square doesn’t. What other everyday shape than a circle has constant width? What is the advantage of that shape having constant width but not being circular?

G2. How could you make a drill bit to drill (almost) square holes?

G3. Three circles of radius 1 are packed together in triangular shape. What is the area in the middle, between them?

G4. A large area is covered by circles packed together triangularly. What percentage of the space is outside the circles?

G5. How many numbers do we need to specify a triangle?

G6. How many numbers do we need to specify a quadrilateral?

G7. How many numbers do we need to specify an n-agon?

G8. You have fencing which comes in 17 hinged sections. How many diagonal lengths do you have to specify to decide exactly what shape that fencing can enclose?

G9. Prove that if a right-angled triangle has whole-number sides, then the product of those whole numbers is divisible by 60 (i.e. the pattern you can see from 3,4,5; 5,12,13; 8,15,17… continues forever).

G10. Prove that *if* a triangle with sides a, b, and c has c^{2}=a^{2}+b^{2}, *then* the triangle is right-angled.

G11. Make a good guess for a rule to find the maximum number of distinct regions that can be made if you mark n points on the circumference of a circle and draw lines to join all the points. (Hint: Look at Pascal’s triangle).

G12. Prove the rule is nC4 + nC2 + 2

**Answers**

N1: 77×87 – 76×87 = (77–76)×87 = 1×87 = 87

N2: 101^{2}-99^{2} = (101-99)×(101+99) = 400. (Neither this, nor N1, as familiar to Year 12 Further Maths students as you’d think).

N3: Can be proved by (strong) induction. Or, better:

N = a + 10b + 100c + 1000d + …

= a + (b + 9b) + (c + 99c) + (d + 999d) + …

= (a + b + c + d + …) + (9b + 99c + 999d + …)

= (a + b + c + d + …) + 9*(b + 11c + 111d + …)

N = (sum of digits of N) + 9 * (some number)

N4: Six. Let b=the units digit of N, so N^{2} = (10a+b)^{2} = 100a^{2}+20ab+b^{2}

So the tens digit of N^{2} is odd only if the tens digit of b^{2} is odd.

But among the numbers 1, 2,…9, only two have squares with odd tens digits, 4 and 6. And in both those cases the units digit of the square is 6.

N5 (borrowed from Terry Tao). If there are s sheep, and each brother has had n×10 roubles before they come to the last share-out, and the younger brother gets a roubles in the last share-out,

then

s^{2} = (2n+1)×10 + a

So, by N4, a=6. The knife is worth 2 roubles.

N5. It makes the angles of an equilateral triangle 66⅔ degrees. 360 has the advantage of more divisibility.

N6. Longer until the weekend.

N7. It was originally 12 hours in a day, when all clocks depended on the sun, so there were no hours when it was dark. The 12 seems to have come from 10 (as in usual base-10) plus one hour for dawn and one for dusk. Originally hours were longer in summer than in winter (though there was not much difference around the Mediterranean, where this scheme originated).

N8. You need to check only prime numbers because if a number is divisible by a composite number, then it is divisible by a smaller prime number too. You need to check primes only up to 7, because if any prime ≥ 11 divides a number ≤ 100, then that prime ≥ 11 must go into the number ≤ 10 times, so there is also a prime ≤ 10 which divides the number.

N9. 2, 3, 5, 7, 11, 13, 17 – all the primes less than the square root of 300.

N10.

No. There will always be at least four students in the right order. First, forget about students and heights: label each student with their number in the height order, so we get a sequence of numbers, e.g. 7 4 8 6 9 0 3 1 5 2 for the picture shown.

Let x(i) = the i’th number, a(i) = length of the biggest ascending sequence including x(i), and d(i) = length of the biggest descending sequence including x(i). [For this problem we start counting from 0, so in the example x(0)=7, a(0)=3, d(0)=4].

Then each number x(i) has a unique combination of a(i) and d(i), in other words if for some j a(i)=a(j) and d(i)=d(j) then i=j.

Why? Suppose j>i. Then either x(j)>x(i), and so a(j) is at least a(i)+1; or x(j)i; by a similar argument it’s impossible that i>j.

If there is no sequence of four students in the right order, then for all i the only possible values of a(i) and d(i) are 1, 2, and 3. There only 9, i.e. 3×3, possible combinations of a(i) and d(i), i.e. only 9 possible values of i. But there are ten numbers in the sequence. So that’s impossible. There must be a sequence of at least four students in the right order.

M1: 1. If the longer sides of the rectangle are 1+d, the smaller are 1-d. Area = (1+d)(1–d) = 1^{2} − d^{2}

M2: √3. Equilateral triangle. Either guessing that the maximum-area triangle is equilateral, or trial and error by drawing triangles, are ok methods of getting there. Better to prove as in M3.

M3: First prove that of all triangles with the same perimeter on the same base, the isosceles triangle has the greatest area. We do it by proving that any non-isosceles triangle with the same area as the isosceles must have a greater perimeter.

ABC is the isoceles triangle on the given base AC, and ADC is another triangle on the same base, i.e. with the same height relative to AC. Construct BC’ as a continuation of AB with length equal to BC.

Then angle DBC = angle ACB (alternate) = angle CAB (isosceles) = angle DBC’ (corresponding)

So triangles DBC and DBC’ are congruent, so DC’ = DC

AD + DC’ > AC’ (two sides of triangle add up to more than the third), so AD + DC > AB + BC. ▇

Now suppose we have a non-equilateral triangle of maximum area with given perimeter.

At least two sides are unequal. But then there is an isosceles triangle of the same perimeter, based on the third side, with greater area than this non-equilateral triangle, which is a contradiction. Therefore the triangle of maximum area with given perimeter is equilateral.

M4. Area.

M5. I don’t know, despite puzzling about it a bit. Something to do with invariants, I suspect.

M6. Four. Let 0.5m=1 unit, then the floor is approx a 6-unit square, so has area a bit less than 36. Or this could be done by drawing the floor on squared paper. Strictly speaking, we don’t know the angles, so the floor might be a long, thin diamond-shape and of very small area. But then you’d have worse problems than some carpet tiles left over. The interesting follow-up is to see that the divergence from area 36 by two sides being 6+6Δ and 6– 6Δ is of order Δ2 (the “top” vertex of the half-floor triangle containing those sides moves along an ellipse as Δ varies), and the divergence from area 36 by the floor’s angles being 90 + Δ and 90 – Δ is also of order Δ2. The area is within 10% of 36 even if Δ = 26 degrees.

M7. The diagonal. Then we could use Heron’s formula or draw an exact diagram on squared paper.

M8. Tie twelve knots on the rope, equally spaced, and make a 3-4-5 triangle.

M9. Because weight increases by the cube of an animal’s size and area (e.g. of cross-section of leg) only by the square. For the same reason, Arctic animals are bigger than animals that live in hot climates, giants would break their ankles every time they walked, large birds can’t fly. (Even eagles, apparently, cannot strictly speaking fly, but only glide on warm-air currents).

P1. Choosing any one particular number if you choose a number genuinely at random from the infinitely many possibilities between, say, 0 and 1. This is of some practical importance. For example, also, with the normal distribution, the probability of the random variable taking any particular precise value is zero. With the Collatz conjecture – “take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1” – it can be proved that the probability of eventually getting back down to 1 is 1, but the conjecture could still be untrue for infinitely many n.

P2. (2n − 1)(2n − 3)….5.3.1, or (2n − 1)!/(n − 1)!2^{(n−1)}

List the competitors in alphabetical order. We have (2n–1) choices for who’ll play the first competitor, (2n–3) for who’ll play the second, and so on down to 1 choice for the last pair.

P3. Yes. P(prize in box you first picked)=⅓. P(prize in box host opened)=0. P(prize in other box)=⅔.

P4. 15 = 6C2. (You can do it just by drawing and counting).

P5. 15, as in P4.

P6. 15, as in P4.

P7. 465 = 31C2.

P8. 27. Reckon “no ice cream” as an option, then we can consider only triple-scoop possibilities. With N flavours plus “no ice cream” we get (N+1)C3 − 1 options. If (N+1)C3 − 1 > 3000, then (N+1)N(N–1)>18006, so N is a bit more than the cube root of 18006. Check out 26 and 27 for N.

P9. About 51% = 1 – 365×364×….×343/36523 = 1–0.4927.

P10. 2.65×1032, or 30! If these seating plans could be reviewed at the rate of 1 per second, 24 hours a day, 365 days a week, it would take about 10^{25} years to review, a time as much longer than the life of the Universe than the life of the Universe is longer than a nanosecond.

P11. 65% = 1 – 0.910

P12. Suppose for simplicity that before the shuffle the deck was the spades in order from ace through king to 2, followed by the hearts in order, the clubs in order, and the diamonds in order, so the first card was the ace of spades. Find the ace of spades in the shuffled deck and swap it with the card now (after the shuffle) in the first position. Find the king of spades and swap it with the card now (after the shuffle) in the second position. And so on. That produces a series of swaps which reverses the shuffle. Do those same swaps in reverse order, and that’s a series of swaps which equals the shuffle.

To prove the shuffle must be even or odd, start by proving that any series of swaps which brings the cards back to exactly the same order as before must be an even number of swaps.

Suppose a friend is doing that series of swaps. You mimic them with a slight change. When she does her first swap, say the first card, the ace of spades, for the three of diamonds, card no.51, you make no swap. You leave those cards where they are, but put a tag on the 51st place in the deck.

From then on, when she swaps her first card with another, you swap the card in the tagged place with that other. When she swaps the card in the tagged place (in her deck) with the card in another position, you swap cards in the same two places, but move the tag to that other place. When she makes any move not involving the first or the tagged places, you mimic exactly.

In that way, your tagged place will always hold the card she has in her first place, her tagged place will always hold her ace of spades, and your first place will always hold your ace of spades.

At some point she has to swap the ace of spades (from her tagged place) back to first place, and to do that put the card she has in her first place into the tagged place. At that point you make no swap, because you already have the ace of spades in our first place, and your friend’s first-place card in your tagged place.

In the end you both get the deck of cards back to the same order, but you’ve done it in two swaps less.

So any series of swaps which brings the deck back exactly as it was can be reduced in length by two. That means the series must include an even number of swaps. Otherwise it could be reduced, in stages, by two swaps each time, until it was one swap long. But a single swap cannot leave the deck exactly as it was.

Now take any permutation, and two chains of swaps s_{1}s_{2}…..s_{n} and t_{1}t_{2}…..t_{m} which are equivalent to it. If we apply the s-swaps, and then the t-swaps in reverse order, we must get back to the original order. Therefore n+m, the length of the total chain of s-swaps and then reversed t-swaps, must be even. That means that if n is even, m must be even; if n is odd, m must be odd. Which is what we wanted to prove ▇

A1. (Borrowed from Terry Tao). Tie your shoelace on the walkway. If the distance is d, the speed of the walkway is u and your walking speed is v, and it takes t seconds to tie your laces, then your journey time tying-off-walkway is [d/(u+v) + t] and tying-on-walkway is [(d–ut)/(u+v) + ut/u]. Alternatively: if A gets on walkway, while B first stops to tie laces, then A gets (u+v)t ahead of B. B reduces that lead by only vt when A stops to tie laces.

A2. 10^{19}. More generally sum from r=1 to r=m of r^{p} = [1/(p+1)]. m^{(p+1)} + terms of lower order

(r+1) r (r–1) (r–2)… (r–8)−r (r–1) (r–2) ….(r–9) = 10 r (r–1) (r–2)… (r–8)

So, summing from r=1 to r= 100, sum for r=1 to r=100 of r^{9} = 1/10 × 100^{10} + terms of order 10^{9} and lower

This general knowledge is probably more important than the specific formulas for sums of r, r<sup2, and r^{3}

10^{9}. More generally sum from= + terms of lower order.

(N+1) N (N–1) (N–2)… (N–8) – N (N–1) (N–2) ….(N–9) = 10 N (N–1) (N–2)… (N–8)

So, summing from N=1 to N=k, 10 + terms of order 109 and lower = 1010 + terms of order 109 and lower

This general knowledge is probably more important than the specific formulas for sums of r, r^{2}, and r^{3}

A3. 10157

ln (n!) = sum from x=1 to x=n of ln(x)

≈integral from 1 to n of ln(x)

≈[x.ln(x)-x] evaluated between 1 and n

= n.ln(n)−n+1

Therefore n! is approx exp [n ln(n) – n + 1]

This is a very crude version of Stirling’s approximation, but tells us for example that 100! (which Casio scientific calculators won’t calculate) is of the order of 37^{100}, so 10^{100}.3.7^{100}, so 10^{157}.

A4. Prove that e is irrational.

Write out e = an infinite series

For any n, multiply both sides by n!

Then n!e = a whole number (first n terms of the series, multiplied by n!) + a remainder

Remainder = 1/(n+1) + 1/(n+1)(n+2) + 1/(n+1)(n+2)(n+3) + ….

Remainder > 0, and remainder < 1/(n+1) + 1/(n+1)^{2} + 1/(n+1)^{3} + ….

So remainder < [1/(n+1)]/[1 − 1/(n+1)] = 1/n < 1

So n!e is not a whole number for any n

e cannot be rational, because it if were rational it would have a denominator N, and N!e would be a whole number.

C1. Zero, since |f(h)|≤h^{2} and thus |[f(h)–f(0)]/h|≤h. Shows that the line of slope f'(a) may cross the curve y=f(x) infinitely many times in any neighbourhood of a.

T1. G, I (if written without serifs), J (ditto), L, M, N, O, P, Q, R, S, U, V, W, Z can.

E, F, H, K, T, X, Y can’t, because they have more than two nodes with an odd number of lines meeting.

T2. No. Too many odd nodes, as above.

G1. The shape of 50p and 20p coins. Having constant width suits them to slot machines, being not circular makes them easier for blind people to identify.

G2. With a drill bit based on the simplest constant-width non-circle shape, the Reuleaux triangle.

http://www.youtube.com/watch?v=L5AzbDJ7KYI

G3. √3−½π

Area of triangle made by connecting the centres of the three circles = √3

Area of inside-circles bits within that triangle = ½π

G4. 9.3%. The entire space is made up of “circle-centre-joining” triangles, so the percentage of the total space which is “outside” = percentage of space inside a “circle-centre-joining” triangle which is “outside” = √3−½&pi/√3 = 9.3%.

G5. Three (as in the congruence cases).

G6. Five (four sides and a diagonal, or four sides and an angle, etc.)

G7. 2n−3. You prove this by induction.

G8. 14 = (2×17–3) – 17

G9. Consider the equation x^{2}+y^{2}=z^{2} in modular (“clock”) arithmetic mod 3, mod 4, and mod 5. There are not enough square numbers mod 3, mod 4, or mod 5 for the equation to have a solution without one of the square numbers being 0, i.e. without x or y or z being divisible by 3, by 4, by 5. This is a generalisation of the easier-to-see fact that at least one of x, y, and z must be even.

G10. From a given point X, draw two lines of lengths a and b at right angles to each other. Then draw another line joining the ends of those lines opposite to X, to form a triangle T. By Pythagoras, that third line has length c defined by c^{2} = a^{2}+b^{2} and c>0.

Let S be any other triangle with sides a, b, and c. S is congruent to T (SSS case). Therefore S is right-angled.

G11. It’s the sum of the first 4 elements of (n−1) row of Pascal’s triangle.

G12. Adapt the proof from lower vertices of the maximum number of areas into which n lines divide the plane being (n^{2}+n+2)/2.

Adapt the circle so that one of the points is at the bottom of the page and the other points are either equally spaced or (if there is an even number of points, in which case even spacing of the points would make all the lines connecting them meet in the centre) just a little different from equally spaced.

Call the lines connecting two marked points on the circumference cross-lines. Then there are nC4 meeting points of cross-lines. Each is a lowest vertex of one region.

We have to add in the regions whose lowest points are meeting points of cross-lines and the circumference.

There are n of those with the chosen bottom point as lowest point.

As you go up the circumference to the right, there are (n − 2), (n − 4)… additional regions at successive points.

As you go up to the left, there are (n − 3), (n − 5)… additional regions.

Total additional regions = n + (n − 2) + (n − 3) + …. + 1 = n (n+1) /2 − (n − 1) = (n^{2} − n)/2 + 1 = nC2 + 1

Therefore total regions = nC4 + nC2 + 1

Since nC2 = (n−1)C1 + (n−1)C2 and nC4 = (n−1)C3 + (n−1)C4 and (n−1)C0 = 1, also

total regions = sum from r=0 to r=4 of (n−1)Cr = first 4 elements of (n−1) row of Pascal triangle, which explains why total regions = 2^{(n−1)} for n<6.

# Some ideas on problem-solving methods in maths

**“The main error to be avoided is trying to attack the problem head-on”** – Alain Connes

**“Face problems with a minimum of blind calculation, a maximum of seeing thought”** – Hermann Minkowski Continue reading

# Report on 9 December prize question

No-one got the prize this fortnight. Daniel Huang discovered how to solve the problem, but didn’t write up his answer. Often mathematical problems are best done by looking at simple cases first, and then doing the more complicated working necessary for the general case. This is an example where the general case is actually easier than the simple cases.