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Monday, 30 June 2014

Sunday Afternoon Maths XIX Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Dartboard

$$\pi (1)^2 - \pi (\frac{1}{2})^2 + \pi (\frac{1}{3})^2 - \pi (\frac{1}{4})^2 + \pi (\frac{1}{5})^2 - ...$$ $$=\sum_{i=1}^\infty \frac{\pi (-1)^{i-1}}{i^2}$$ $$=\pi\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}$$ $$=\pi\left(\frac{\pi^2}{12}\right)$$ $$=\frac{\pi^3}{12}$$
Extension
Prove that
$$=\sum_{i=1}^\infty \frac{(-1)^{i-1}}{i^2}=\frac{\pi^2}{12}$$

Multiples of Three

If $$10A+B=3n$$, then:
$$100A+B = 100A-10A+10A+B$$$$=90A+3n$$$$=3(30A+n)$$
So $$A0B\div3=30A+n$$.
Extension
What are $$A00B\div3$$, $$A000B\div3$$, $$A0000B\div3$$, etc?

Sunday, 29 June 2014

Sunday Afternoon Maths XIX

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

Dartboard

Concentric circles with radii 1, $$\frac{1}{2}$$, $$\frac{1}{3}$$, $$\frac{1}{4}$$, ... are drawn. Alternate donut-shaped regions are shaded.
What is the total shaded area?

Multiples of Three

If the digits of a number add up to a multiple of three, then the number is a multiple of three. Therefore if a two digit number, $$AB$$ (first digit $$A$$, second digit $$B$$; not $$A\times B$$), is a multiple of three, then $$A0B$$ is also a multiple of three.
If $$AB\div 3=n$$, then what is $$A0B\div 3$$?

Monday, 23 June 2014

Sunday Afternoon Maths XVIII Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Parabola

The co-ordinates of the points where the lines intersect the parabola are $$(a,a^2)$$ and $$(-b,b^2)$$. Hence the gradient of the line between them is:
$$\frac{a^2-b^2}{a-(-b)}=\frac{(a+b)(a-b)}{a+b}=a-b$$
Therefore the y-coordinate is:
$$b^2 + b(a-b) = ba$$
Ferdinand Möbius, who discovered this property called the curve a Multiplicationsmaschine or 'multipliction machine' as it could be used to perform multiplication.
Extension
How could you use the graph of $$y=x^2$$ to divide 100 by 7?

Seven Digits

Let's call Dr. Dingo's number $$n$$. If the number is squared twice then multiplied by $$n$$, we get $$n^5$$.
For all integers $$n$$, the final digit of $$n^5$$ is the same as the final digit of $$n$$. In other words:
$$n^5\equiv n \mod 10$$
Therefore, the final digit of Dr. Dingo's number is 7.
$$7^5=16807$$ $$17^5=1419857$$ $$27^5=14348907$$
So, in order for the answer to have seven digits, Dr. Dingo's number was 17.
Extension
For which integers $$m$$ does there exist an integer $$n$$ such that for all integers $$x$$:
$$x^n\equiv x \mod m$$

Sunday, 22 June 2014

Sunday Afternoon Maths XVIII

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

Parabola

On a graph of $$y=x^2$$, two lines are drawn at $$x=a$$ and $$x=-b$$ (for $$a,b>0$$. The points where these lines intersect the parabola are connected.
What is the y-coordinate of the point where this line intersects the y-axis?

Seven Digits

"I'm thinking of a number. I've squared it. I've squared the square. And I've multiplied the second square by the original number. So I now have a number of seven digits whose final digit is a 7," said Dr. Dingo to his daughter.
Can you work out Dr. Dingo's number?

Saturday, 21 June 2014

Tennis Maths Winning with the Smallest Share of Points & Comparing Players with Serving Stats

With World Cup fever taking over, you may have forgotten that Wimbledon is just a few days away.

Tennis Scoring

Tennis matches are split into sets (three sets for ladies' matches, five sets for men's), which are in turn split into games. The players take it in turns to serve for a game. The scoring in a game is probably best explained with a flowchart (click to enlarge):
To win a set, a player must win at least six games and two more games than their opponent. If the score reaches six games all, then a tie break is played. In this tie break, the first player to win at least seven points and two points more than their opponent wins. In the final set there is no tie break, so matches can last a long time.

Winning with the Smallest Share of Points

Due to the way tennis is split into sets and games, the player who wins the most points will not necessarily win the match. This got me thinking: what is the smallest proportion of points which can be won while still winning the tennis match?
First, let's consider a men's match. In order to win with the lowest proportion of points, our player should let his opponent win two sets without winning a point and win the other three sets. In the two lost sets, the opponent should win 0-6 taking every point: in total the opponent will win 48 points in these sets.
Leaving the final set for now, the other two sets are won by our player. To win these with the smallest proportion of the points, they should be won 7-6 on a tie break. In the 6 lost games, the opponent should take all the points. In the won games and the tie break, our player should win by two points with the lowest total score. (Winning with more than the lowest total score will mean both players win an equal number of extra points, moving the proportion of points our player wins closer to 50%, higher than it needs to be.)
Therefore, our player will win 4 points out of 6 in the games he wins, win 0 out of 4 points in the games he loses and wins the tie break 7 points to 5. This means that in total our player will 62 points out of 144 in the two won sets.
For the same reason as above, the final set should be won with the lowest total score: 6-4. Using the same scores for each game, our player wins 24 points out of 52.
Overall, our player has won 86 points out of 244, a mere 35% of the points.
If the match is a ladies' match then the same analysis will work, but with each player winning one less set. This gives our player 55 points out of 148, 37% of the points.
This result demonstrates why tennis remains exciting through the whole match. The way tennis is split into sets and games means that our opponent can win 65% of the points but if the pressure gets to them at the most important points, our player can still win the match. This makes for a far more interesting competition than a simple race to one hundred points which could quickly become a foregone conclusion.

Comparing Players with Serving Stats

During tennis matches, players are often compared using statistics such as the percentages of serves which are successful. Imagine a match between Player A and Player B.
In the first set, Player A and Player B are successful with 100% and 92% of their serves respectively. In the second set, these figures are 56% and 48%. Player A clearly looks to be the better server, as they have a higher percentage in each set. However if we look at the two sets in more detail:
 Player A Player B First Set 20/20 67/73 Second Set 45/80 13/27 Total 65/100 80/100
Table showing successful serves/total serves.
Overall, Player B has an 80% serve success rate, while Player A only manages 65%.
This is an example of Simpson's paradox: a trend which appears in the set-by-set data disappears when the data is combined. This occurs because when we look at the set-by-set percentages, the total number of serves is not taken into account: Player A served more in the second set so their overall percentage will be closer to 56%; Player B served more in the first set so their overall percentage will be closer to 92%.

Monday, 16 June 2014

Sunday Afternoon Maths XVII Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Maths Jam

15th February 2022.
Extension
What is the latest date in the month on which Maths Jam can fall and when will this next happen?

N

(b) implies that the digits of $$N$$ are all 1 or 7, so $$N$$ can only be 111, 117, 171, 177, 711, 717, 771 or 777. These are all divisible by 3, so no such integers $$N$$ exist.
Extension
Consider 21-digit integers $$N$$ such that:
(a) $$N$$ is not exactly divisible by 2, 3 or 5.
(b) No digit of $$N$$ is exactly divisible by 2, 3 or 5.
How many such integers $$N$$ are there?

Square Numbers

Let $$a^2$$ and $$b^2$$ be the two square numbers.
$$2(a^2 +b^2 ) = 2a^2 +2b^2$$ $$= a^2 + 2ab + b^2 + a^2 - 2ab + b^2$$ $$= (a+b)^2 +(a-b)^2$$
Extension
Prove that 3 times the sum of 3 squares is also the sum of 4 squares.

Differentiate This

$$f(x)=e^{x^{ \frac{\ln{\left(\ln{x}\right)}}{ \ln{x}}} }$$ $$=e^{e^{ \frac{\ln{\left(\ln{x}\right)}}{ \ln{x}}\ln{x}} }$$ $$=e^{e^{ \ln{\left(\ln{x}\right)}} }$$ $$=e^{\ln{x} }$$ $$=x$$
Therefore:
$$f'(x)=1$$

Sunday, 15 June 2014

Sunday Afternoon Maths XVII

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit. This Tuesday is Maths Jam, make sure you're there!

Maths Jam

Maths Jam is always held on the second-to-last Tuesday of the month. This month, it will be held on the 17th. What is the earliest date in the month on which Maths Jam can fall and when will this next happen?

N

Consider three-digit integers $$N$$ such that:
(a) $$N$$ is not exactly divisible by 2, 3 or 5.
(b) No digit of $$N$$ is exactly divisible by 2, 3 or 5.
How many such integers $$N$$ are there?

Square Numbers

Towards the end of his life, Lewis Carroll recorded in his diary that he had discovered that double the sum of two square numbers could always be written as the sum of two square numbers. For example
$$2(3^2 +4^2 )=1^2 +7^2$$ $$2(5^2 +8^2 )=3^2 +13^2$$
Prove that this can be done for any two square numbers.

Differentiate This

$$f(x)=e^{x^{ \frac{\ln{\left(\ln{x}\right)}}{ \ln{x}}} }$$
Find $$f'(x)$$.

Monday, 9 June 2014

Sunday Afternoon Maths XVI Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Always a Multiple?

Let the two digit number chosen by $$10a+b$$, with $$a$$ and $$b$$ one digit integers. Reversed this will be $$10b+a$$. The sum of these will be $$11a+11b$$ which is divisible by 11.

This will work for any integer with an even number of digits. Let our number have $$2n$$ digits. It can be written as:
$$\sum_{i=1}^{2n}10^{i-1} a_i$$
Adding it to its reverse, we get:
$$\sum_{i=1}^{2n}10^{i-1} a_i + \sum_{i=1}^{2n}10^{2n-i} a_i = \sum_{i=1}^{2n}(10^{i-1}+10^{2n-i}) a_i$$
$$10^{i-1}+10^{2n-i}$$ is divisible by 11 (for $$n \in \mathbb{N}$$, $$i \in \mathbb{N}$$, $$i\leq 2n$$). This can be shown by induction on $$n$$:
If $$n=1$$: $$10^{i-1}+10^{2n-i} = 10^{i-1}+10^{2-i}=11$$, which is clearly divisible by 11.
Suppose result is true for $$n-1$$. Now consider $$10^{i-1}+10^{2n-i}$$.
If $$i>1$$, then $$10^{i-1}+10^{2n-i}= 10(10^{(i-1)-1}+10^{(2n-2)-(i-1)}$$ which is divisible by 11 by the inductive hypothesis.
If $$i=1$$, then:
$$10^{i-1}+10^{2n-i} = 1+10^{2n-1} = \sum_{j=1}^{j=2n-2}9\times 10^j +11$$ $$=\sum_{j=1}^{j=n-1}99\times 10^{2j} +11$$ $$=11\left(\sum_{j=1}^{j=n-1}9\times 10^{2j} +1\right)$$
Extension
Which numbers with an odd number of digits will be divisible by 11 when added to their reverse?

Pocket Money

If Amy gets $$n$$p for pocket money then brother 1 gets $$\frac{n}{2}$$p, brother 2 gets $$\frac{n}{3}$$p, brother 3 gets $$\frac{n}{4}$$p and brother 4 gets $$\frac{n}{5}$$p.
If Tom is brother 1 and Peter is brother 2, then:
$$\frac{n}{2}-\frac{n}{3}=30$$ $$\frac{n}{6}=30$$ $$n=180$$
If Tom is brother 1 and Peter is brother 3, then:
$$\frac{n}{2}-\frac{n}{4}=30$$ $$\frac{n}{4}=30$$ $$n=120$$
If Tom is brother 1 and Peter is brother 4, then:
$$\frac{n}{2}-\frac{n}{5}=30$$ $$\frac{3n}{10}=30$$ $$n=100$$
If Tom is brother 2 and Peter is brother 3, then:
$$\frac{n}{3}-\frac{n}{4}=30$$ $$\frac{n}{12}=30$$ $$n=360$$
If Tom is brother 2 and Peter is brother 4, then:
$$\frac{n}{3}-\frac{n}{5}=30$$ $$\frac{2n}{15}=30$$ $$n=225$$
If Tom is brother 3 and Peter is brother 4, then:
$$\frac{n}{4}-\frac{n}{5}=30$$ $$\frac{n}{20}=30$$ $$n=600$$
So the possible amounts of money Amy could have received are £1.80, £1.20, £1, £3.60, £2.25 and £6.
Extension
Which values could the 30p be replaced with and still give a whole number of pence for all the possible answers?

Sunday, 8 June 2014

Sunday Afternoon Maths XVI

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

Always a Multiple?

Prove that the answer will be a multiple of 11 for any starting number.
Will this work with three digit numbers? Four digit numbers? $$n$$ digit numbers?

Pocket Money

When Dad gave out the pocket money, Amy received twice as much as her first brother, three times as much as the second, four times as much as the third and five times as much as the last brother. Peter complained that he had received 30p less than Tom.
Use this information to find all the possible amounts of money that Amy could have received.

Monday, 2 June 2014

Sunday Afternoon Maths XV Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

Grand Piano

When halfway around the corner, the grand piano will look like this:
The problem then is finding the largest area rectangle which fits in the highlighted triangle: an isosceles triangle where the base is twice the height. Let the base be 2 and the height be 1.
If the height of the rectangle is $$h$$, then its width is $$2(1-h)$$. Therefore its area is $$2h-2h^2$$. By differentiation, it can be seen that this is maximum when $$h=\frac{1}{2}$$, which means that ratio of the rectangle's length to its width is 2:1.
Extension
If the corner was not a 90° angle, then what is the largest area rectangle which could fit round it?

Cycling Digits

A few examples are:
$$105263157894736842 \times 2 = 210526315789473684$$ $$210526315789473684 \times 2 = 421052631578947368$$ $$421052631578947368 \times 2 = 842105263157894736$$
Extension
I have in mind a number which when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 3. Am I telling the truth?

The Mutilated Chessboard

On a normal chessboard there are 32 white and 32 black squares. After removing two diagonally opposite corners there will be 30 white and 32 black or 32 black and 30 white squares.
Each domino covers one white square and one black square. Therefore a combination of dominoes will always cover the same number of black and white squares, so it is not possible to cover all the squares.
Extension
Is it possible to do a knight's tour on the mutilated chessboard?

Tennis

A set can be won by hitting the ball 12 times: three service games can be won with four aces, the opponent loses their service games with four double faults.

Sunday, 1 June 2014

Sunday Afternoon Maths XV

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.

Grand Piano

Jack and Jill are moving into a new flat and their grand piano presents a potential problem. Fortunately, it will just pass round the corridor without being tipped or disassembled.
Given that its area, looking down from above, is the largest possible which can be passed around the corner, what is the ratio of its length to its width?

Cycling Digits

I have in mind a number which when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 2. Am I telling the truth?

The Mutilated Chessboard

You are given a chessboard where two diagonally opposite corners have been removed and a large bag of dominoes of such size that they exactly cover two adjacent squares on the chessboard.
Is it possible to place 31 dominoes on the chessboard so that all the squares are covered? If yes, how? If no, why not?

Tennis

What is the minimum number of times a player has to hit the ball in a set of tennis and win a standard set (the set is not ended by injury, disqualification, etc.)?