Monday, 25 August 2014

Sunday Afternoon Maths XXVI Answers & Extensions

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

Twenty-One

Virgil should go second. Whatever Scott adds, Virgil should then add to make four. For example, if Scott says 3, Virgil should say 1.
Using this strategy, Virgil will say 4, 8, 12, 16 then 20, forcing Scott to go above 21.
Extension
(i) If instead of 21, 22 cannot be said/beaten, how should Virgil win? How about 23? Or 24? How about \(n\)?
(ii) If instead of adding 1 to 3, 1 to 4 can be added, how should Virgil win? How about 1 to 5? Or 2 to 5? How about \(m\) to \(l\)?
(iii) Alan wants to join the game. Can Virgil win if there are three people? Can he win if there are \(k\) people?

Odd and Even Outputs

\(n\)
oddeven
\(m\)oddoddodd
eoddodd
\(g(n,m)=1\)

\(n\)
oddeven
\(m\)oddoddodd
eoddeven
\(g(n,m)=n\times m + n + m\)

\(n\)
oddeven
\(m\)oddoddodd
eevenodd
\(g(n,m)=n\times m +n+1\)

\(n\)
oddeven
\(m\)oddoddodd
eeveneven
\(g(n,m)=m\)
\(n\)
oddeven
\(m\)oddoddeven
eoddodd
\(g(n,m)=n\times m+m+1\)

\(n\)
oddeven
\(m\)oddoddeven
eoddeven
\(g(n,m)=n\)

\(n\)
oddeven
\(m\)oddoddeven
eevenodd
\(g(n,m)=n+m+1\)

\(n\)
oddeven
\(m\)oddoddeven
eeveneven
\(g(n,m)=n\times m\)
\(n\)
oddeven
\(m\)oddevenodd
eoddodd
\(g(n,m)=n\times m+1\)

\(n\)
oddeven
\(m\)oddevenodd
eoddeven
\(g(n,m)=n+m\)

\(n\)
oddeven
\(m\)oddevenodd
eevenodd
\(g(n,m)=n+1\)

\(n\)
oddeven
\(m\)oddevenodd
eeveneven
\(g(n,m)=n\times m+n\)
\(n\)
oddeven
\(m\)oddeveneven
eoddodd
\(g(n,m)=m+1\)

\(n\)
oddeven
\(m\)oddeveneven
eoddeven
\(g(n,m)=n\times m+n\)

\(n\)
oddeven
\(m\)oddeveneven
eevenodd
\(g(n,m)=n\times m+n+m+1\)

\(n\)
oddeven
\(m\)oddeveneven
eeveneven
\(g(n,m)=2\)
Extension
Can you find functions \(h:\mathbb{N}\times\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}\) (call the inputs \(n\), \(m\) and \(l\)) to give the following outputs:
\(l\) odd
\(n\)
oddeven
\(m\)oddeveneven
eeveneven
\(l\) even
\(n\)
oddeven
\(m\)oddeveneven
eeveneven

\(l\) odd
\(n\)
oddeven
\(m\)oddeveneven
eeveneven
\(l\) even
\(n\)
oddeven
\(m\)oddeveneven
eevenodd

etc

Sunday, 24 August 2014

Sunday Afternoon Maths XXVI

Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit.
It's no late to get tickets for EMF Camp, where I will be giving a talk on flexagons, folding tube maps and braiding.

Twenty-One

Scott and Virgil are playing a game. In the game the first player says 1, 2 or 3, then the next player can add 1, 2 or 3 to the number and so on. The player who is forced to say 21 or above loses. The first game went like so:
Scott: 3
Virgil: 4
Scott: 5
Virgil: 6
Scott: 9
Virgil: 12
Scott: 15
Virgil 17
Scott: 20
Virgil: 21
Virgil loses.
To give him a better chance of winning, Scott lets Virgil choose whether to go first or second in the next game. What should Virgil do?

Odd and Even Outputs

Let \(g:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}\) be a function.
This means that \(g\) takes two natural number inputs and gives one natural number output. For example if \(g\) is defined by \(g(n,m)=n+m\) then \(g(3,4)=7\) and \(g(10,2)=12\).
The function \(g(n,m)=n+m\) will give an even output if \(n\) and \(m\) are both odd or both even and an odd output if one is odd and the other is even. This could be summarised in the following table:
\(n\)
oddeven
\(m\)oddevenodd
eoddeven
Using only \(+\) and \(\times\), can you construct functions \(g(n,m)\) which give the following output tables:
\(n\)
oddeven
\(m\)oddoddodd
eoddodd
\(n\)
oddeven
\(m\)oddoddodd
eoddeven
\(n\)
oddeven
\(m\)oddoddodd
eevenodd
\(n\)
oddeven
\(m\)oddoddodd
eeveneven
\(n\)
oddeven
\(m\)oddoddeven
eoddodd
\(n\)
oddeven
\(m\)oddoddeven
eoddeven
\(n\)
oddeven
\(m\)oddoddeven
eevenodd
\(n\)
oddeven
\(m\)oddoddeven
eeveneven
\(n\)
oddeven
\(m\)oddevenodd
eoddodd
\(n\)
oddeven
\(m\)oddevenodd
eoddeven
\(n\)
oddeven
\(m\)oddevenodd
eevenodd
\(n\)
oddeven
\(m\)oddevenodd
eeveneven
\(n\)
oddeven
\(m\)oddeveneven
eoddodd
\(n\)
oddeven
\(m\)oddeveneven
eoddeven
\(n\)
oddeven
\(m\)oddeveneven
eevenodd
\(n\)
oddeven
\(m\)oddeveneven
eeveneven

Monday, 18 August 2014

Sunday Afternoon Maths XXV Answers & Extensions

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

Whist

Represent each player by a compass point. Let B, D, A and S represent Messrs. Banker, Dentist, Apothecary and Scrivener respectively and b, d, a and s represent the four jobs.
Mr. Banker (B) partners the scrivener (s). Let B sit at West. This means s sits at East. As no other player can also be B or s, the table looks like this:
North
BDAS
bdas
West
BDAS
bdas
East
BDAS
bdas
South
BDAS
bdas
The dentist (d) sits on Mr. Scrivener's (S) right. East cannot be S, so North cannot be d. East cannot be d, so South cannot be S.
North
BDAS
bdas
West
BDAS
bdas
East
BDAS
bdas
South
BDAS
bdas
By elimination, only North can be S. This means that d must sit to the right of North (at West):
North
BDAS
bdas
West
BDAS
bdas
East
BDAS
bdas
South
BDAS
bdas
A and a are partners. This is only possible if A is South and a is North:
North
BDAS
bdas
West
BDAS
bdas
East
BDAS
bdas
South
BDAS
bdas
Therefore, Mr. Banker the dentist sits to the left of the banker.
Extension
If each person is partnered with their job namesake, how many possible combinations of names and jobs are possible?

Polya Strikes Out

1, 3, 7, 12, 19, ...
1, 7, 19, ...
1=1; 1+7=8; 1+7+19=27; ...
1, 8, 27, ...
The final sequence is the cube numbers. To show why, let \(n\) be an integer and follow through the process.
Cross out every third number:
1, 2, 3, 4, 5, 6, ..., 3n, \(3n+1\), \(3n+2\), ...
1, 2, 4, 5, ..., \(3n+1\), \(3n+2\), ...
Find the cumulative sums:
$$1=1$$ $$1+2=1+2=3$$ $$1+2+4=1+2+3+4-3=7$$ $$1+2+4+5=1+2+3+4+5-3=12$$ $$...$$ $$1+2+4+5+...+(3n+1)=\sum_{i=1}^{3n+1}-\sum_{i=1}^{n}3i$$ $$=\frac{1}{2}(3n+1)(3n+2)-\frac{3}{2}n(n+1)$$ $$=3n^2+3n+1$$ $$1+2+4+5+...+(3n+2)=3n^2+3n+1+(3n+2)$$ $$=3n^2+6n+3$$ $$...$$
1, 3, 7, 12, ..., \(3n^2+3n+1\), \(3n^2+6n+3\), ...
Cross out every second number, starting with the second:
1, 3, 7, 12, ..., \(3n^2+3n+1\), 3n2+6n+3, ...
1, 7, ..., \(3n^2+3n+1\), ...
Find the cumulative sums. The \(m\)th sum is:
$$\sum_{n=0}^{m}3n^2+3n+1$$ $$=3\sum_{n=0}^{m}n^2+3\sum_{n=0}^{m}n+\sum_{n=0}^{m}1$$ $$=\frac{3}{6}m(m+1)(2m+1)+\frac{3}{2}m(m+1)+m+1$$ $$=\frac{1}{2}(m+1)(m(2m+1)+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+m+3m+2)$$ $$=\frac{1}{2}(m+1)(2m^2+4m+2)$$ $$=(m+1)(m^2+2m+1)$$ $$=(m+1)(m+1)^2$$ $$=(m+1)^3$$
Hence the numbers obtained are the cube numbers.
Extension
What happens if you cross out every third number starting at the second? Or every fifth number starting at the fifth? Or every \(n\)th number starting at the \(m\)th?

Sunday, 17 August 2014

Sunday Afternoon Maths XXV

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!
It's less than two weeks 'til EMF Camp, where there will be talks on maths and science and a whole load of other awesome stuff. Tickets here

Whist

Messrs. Banker, Dentist, Apothecary and Scrivener played whist last night. (whist is a four player card game where partners sit opposite each other.) Each of these gentlemen is the namesake of another's vocation.
Last night, the apothecary partnered Mr. Apothecary; Mr. Banker's partner was the scrivener; on Mr. Scrivener's right sat the dentist.
Who sat on the banker's left?

Polya Strikes Out

Write the numbers 1, 2, 3, ... in a row. Strike out every third number beginning with the third. Write down the cumulative sums of what remains:
1, 2, 3, 4, 5, 6, 7, ...
1, 2, 3, 4, 5, 6, 7, ...
1, 2, 4, 5, 7, ...
1=1; 1+2=3; 1+2+4=7; 1+2+4+5=12; 1+2+4+5+7=19; ...
1, 3, 7, 12, 19, ...
Now strike out every second number beginning with the second. Write down the cumulative sums of what remains. What is the final sequence? Why do you get this sequence?

Monday, 11 August 2014

Sunday Afternoon Maths XXIV Answers & Extensions

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

Square Cross

Label and orient the shape as follows:
The smaller diagonal square is made up of a 1×1 square and four 1×2 right-angled triangles. Therefore its area is 5.
The line FM, and therefore the line DN, has gradient 2. The line JM, and therefore the line ON, has gradient -½. ON passes through (1,2) and DN passes through (2,0). Therefore, DN has equation \(y=2x-4\) and ON has equation \(y=\frac{5}{2}-\frac{1}{2}x\). These lines intersect at \((\frac{13}{5},\frac{6}{5})\), these are the co-ordinates of N. By the same method, the co-ordinates of P are \((-\frac{8}{5},-\frac{1}{5})\).
By Pythagoras' Theorem, The diagonal of the larger square is \(\frac{7\sqrt{10}}{5}\) and so the area of the larger square is 9.8.

Exact Change

In the UK, eight coins are needed: 1p, 1p, 2p, 5p, 10p, 20p, 20p, 50p.
In the US, ten coins are needed: 1¢, 1¢, 1¢, 1¢, 5¢, 10¢, 10¢, 25¢, 25¢, 25¢.
Extension
In a far away country, the unit of currency is the #, which is split into 100@ (# is like £ or $; @ is like p or ¢).
Let C be the number of coins less than #1. Let P be the number of coins needed to make any value between 1@ and 99@. Which coins should be the country mint to minimise the value of P+C?

Sunday, 10 August 2014

Sunday Afternoon Maths XXIV

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

Square Cross

A figure in the shape of a cross is made from five 1 x 1 squares, as shown. The cross is inscribed in a large square whose sides are parallel to the dashed square, formed by four vertices of the cross.
What is the area of the large outer square?

Exact Change

In the UK, the coins less than £1 are 1p, 2p, 5p, 10p, 20p and 50p. How many coins would I need to carry in my pocket so that I could make any value from 1p to 99p?
In the US, the coins less than $1 are 1¢, 5¢, 10¢, 25¢. How many coins would I need to carry in my pocket so that I could make any value from 1¢ to 99¢?

Monday, 4 August 2014

Sunday Afternoon Maths XXIII Answers & Extensions

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

Let the Passenger Train Through!

The goods train reverses into the siding, leaving three trucks, then reverses out of the way.
The passenger train attaches the three trucks to its front then reverses out of the way.
The goods train backs into the siding.
The passenger train drives past the siding.
The goods train drives forwards out of the way.
The passenger train reverses, then drives forward into the siding, leaving the three trucks. It then drives forward out of the way.
The goods train reverses into the siding, picking up the trucks. The trains have now passed each other.
Extension
1. If the goods train can drive forwards into the siding, how could it let the passenger train through?
2. If the goods train had 8 trucks, how could it let the passenger train through?

Cooked Turkey

Let the first digit be \(a\) and the final digit be \(b\). The cost of the 72 turkeys (in pence) is \(10000a+6790+b\). This must be divisible by 72.
$$10000a+6790+b=(138\times 72+64)a+(94\times 72+22)+b$$ $$=(138a+94)\times 72+64a+22+b$$
\(10000a+6790+b\) is divisible by 72, so \(64a+22+b\) is divisible by 72. This means that \(b\) must be even (as 64,22,72 are all even). Let \(b=2c\).
Dividing by two, we find that \(32a+11+c\) is divisible by 36. \(c\) must be odd, so that \(32a+11+c\) is even. Let \(c=2d+1\) and so \(b=4d+2\).
Dividing by two again, we find that \(16a+6+d\) is divisible by 18. \(d\) must be even, so that \(16a+6+d\) is even. Let \(d=2e\) and so \(b=8e+2\). But \(b\) is a single digit number, so \(e=0\), \(b=2\) and \(16a+6\) is divisible by 18.
Dividing by two yet again, we find that \(8a+3\) is divisible by 9. \(a\) must be divisible by 3. Let \(a=3f\).
Dividing by three, we find that \(8f+1\) is divisible by 3. \(8f+1=6f+2f+1\) so \(2f+1\) is divisible by 3. This will be true when \(f=1,4,7,10,...\). \(a\) must be a single digit, so \(f=1\) and \(a=3\). And so the price of the 72 turkeys is £367.92, and one turkey will cost £5.11.
Extension
Which numbers could 72 be replaced with in the original problem so that the problem still has a unique solution?

Mrs. Coldcream

Let \(G\) be the number of girls, and £\(x\) be the amount each girl was originally to receive. This means that there are \(19-G\) boys who were to receive £\(x+30\). The total scholarship is £1000, so:
$$xG+(x+30)(19-G)=1000$$
This simplifies to:
$$19x-30G=430$$
\(G\) is less than 19, so the only integer solution of this equation is \(x=40\) and \(G=11\).
After the funds are redistributed, each girl therefore receives £48. Let \(B\) be the amount each boy will now receive. The total is still £1000, so:
$$48\times 11+8B=1000$$
Which can be solved to find \(B=59\).
Each girl receives £48 and each boy receives £59.
Extension
Mrs. Coldcream still feels that this is unfair. How much more should be added to each girl's scholarship (resizing the boys' accordingly) so that the girls get at least as much as the boys?

Ten Digit Number

6210001000 has 6 zeros, 2 ones, 1 two, 0 threes, 0 fours, 0 fives, 1 six, 0 sevens, 0 eights and 0 nines.
Extension
Are there any more numbers like this? Prove that you have them all.