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.
(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?
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | odd | odd |
\(g(n,m)=m+1\)
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | odd | even |
\(g(n,m)=n\times m+n\)
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | even | odd |
\(g(n,m)=n\times m+n+m+1\)
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | even | even |
\(g(n,m)=2\)
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
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | even | even |
|
\(l\) even
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | even | even |
|
\(l\) odd
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | even | even |
|
\(l\) even
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | even | odd |
|
etc
Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit. 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:
| \(n\) |
| odd | even |
| \(m\) | odd | even | odd |
| e | odd | even |
Using only \(+\) and \(\times\), can you construct functions \(g(n,m)\) which give the following output tables:
| \(n\) |
| odd | even |
| \(m\) | odd | odd | odd |
| e | odd | odd |
|
| \(n\) |
| odd | even |
| \(m\) | odd | odd | odd |
| e | odd | even |
|
| \(n\) |
| odd | even |
| \(m\) | odd | odd | odd |
| e | even | odd |
|
| \(n\) |
| odd | even |
| \(m\) | odd | odd | odd |
| e | even | even |
|
| \(n\) |
| odd | even |
| \(m\) | odd | odd | even |
| e | odd | odd |
|
| \(n\) |
| odd | even |
| \(m\) | odd | odd | even |
| e | odd | even |
|
| \(n\) |
| odd | even |
| \(m\) | odd | odd | even |
| e | even | odd |
|
| \(n\) |
| odd | even |
| \(m\) | odd | odd | even |
| e | even | even |
|
| \(n\) |
| odd | even |
| \(m\) | odd | even | odd |
| e | odd | odd |
|
| \(n\) |
| odd | even |
| \(m\) | odd | even | odd |
| e | odd | even |
|
| \(n\) |
| odd | even |
| \(m\) | odd | even | odd |
| e | even | odd |
|
| \(n\) |
| odd | even |
| \(m\) | odd | even | odd |
| e | even | even |
|
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | odd | odd |
|
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | odd | even |
|
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | even | odd |
|
| \(n\) |
| odd | even |
| \(m\) | odd | even | even |
| e | even | even |
|
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:
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.
By elimination, only North can be S. This means that d must sit to the right of North (at West):
A and a are partners. This is only possible if A is South and a is North:
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?
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! 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?
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?
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¢?
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.
Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit. Let the Passenger Train Through!
A goods train, made up of a locomotive and 5 trucks, stops at a small station. The small station has a siding which the goods train can reverse into. The siding can hold an engine and two trucks or three trucks.
A passenger train arrives travelling in the opposite direction. How can they let it through? (The passenger train is too long to fit in the siding.)
Cooked Turkey
An old invoice showed that seventy-two turkeys had been purchased for "—67.9—". The first and last digits were illegible.
How much did one turkey cost?
Mrs. Coldcream Objected
"I object," said Councillor Mrs. Coldcream. "I see no reason why the boys should be so favoured at the expense of the girls."
This was at a meeting of the Holmshire Education Committee. It had been proposed to award 19 scholarships totalling £1000 to boys and girls of the county. It had been proposed that each girl receive a set amount and each boy receive £30 more than each girl.
Mrs. Coldcream pressed her point with such fervour that it was decided to reallocate the money. Each girl would receive £8 more than originally proposed, with the boys' scholarships scaled down accordingly.
How much did each boy and each girl receive?
Ten Digit Number
Can you create a 10-digit number, where the first digit is how many zeros in the number, the second digit is how many 1s in the number etc. until the tenth digit which is how many 9s in the number?
Coming and Going
Add up the number of doors leaving each room; call the sum \(S\). As the number in each room is even, \(S\) will be even. Each interior door has been counted twice (as they can be seen in two rooms) and each exterior door has been counted once. Let \(I\) be the number of interior doors and \(E\) be the number of exterior doors. We have:
$$S=2I+E$$$$E=S-2I$$
\(S\) and \(2I\) are even, so \(E\) must be even.
Extension
If the number of doors in each room is odd, is the number of exterior doors odd or even?
Overlapping Triangles
Let \(S\) be the area of the large square, \(T\) be the area of one of the large triangles, \(U\) be one of the red overlaps and \(V\) be the uncovered blue square. We can write:
$$S=4T-4U+V$$
as the area of the square is the total of the four triangles, take away the overlaps as they have been double counted, add the blue square as it has been missed.
We know that \(4U=V\), so:
$$S=4T-V+V$$$$S=4T$$
Therefore one of the triangles covers one quarter of the square.
Extension
Five congruent triangles are drawn in a regular pentagon. The total area which the triangles overlap (red) is equal to the area they don't cover (blue). What proportion of the area of the large pentagon does each triangle take up?
\(n\) congruent triangles are drawn in a regular \(n\) sided polygon. The total area which the triangles overlap is equal to the area they don't cover. What proportion of the area of the large \(n\) sided polygon does each triangle take up?
Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit. Coming and Going
In my house are a number of rooms. (A hall separated from the rest of the house by one or more doors counts as a room.) Each room has an even number of doors, including doors that lead outside. Is the total number of outside doors even or odd?
Overlapping Triangles
Four congruent triangles are drawn in a square.
The total area which the triangles overlap (red) is equal to the area they don't cover (blue). What proportion of the area of the large square does each (purple) triangle take up?
Sum Equals Product
If \(a+b=a\times b\), then:
$$ab-b=a$$
$$b(a-1)=a$$
$$b=\frac{a}{a-1}$$
This will work for any \(a \not= 1\) (\(a=1\) will not work as this will mean division by zero).
Extension
(i) Given a number \(a\), can you find a number \(b\) such that \(b-a =\frac{b}{a}\)?
(ii) Given a number \(a\), can you find a number \(b\) such that \(b-a =\frac{a}{b}\)?
(iii) Given a number \(a\), can you find a number \(b\) such that \(a-b =\frac{b}{a}\)?
(iv) Given a number \(a\), can you find a number \(b\) such that \(a-b =\frac{a}{b}\)?
Wool Circles
Starting with the person who starts with the wool and going anti-clockwise, number the people \(0,1,2,3,4,...\). As the wool is passed, it will be held by people with numbers:
$$0,a+1,2(a+1),3(a+1),...,k(a+1),...$$
For example, if \(n=10\) and \(a=3\):
The first person will have the wool again when:
$$k(a+1)\equiv 0 \mod n$$
or
$$k(a+1)=ln$$
$$l=\frac{a+1}{\mbox{hcf}(a+1,n)} \mbox{ and } k=\frac{n}{\mbox{hcf}(a+1,n)}$$
\(k\) is also the number of people who are holding the wool. So the number of different coloured balls needed is:
$$\frac{n}{\left(\frac{n}{\mbox{hcf}(a+1,n)}\right)}$$$$= \mbox{hcf}(a+1,n)$$
Extension
The ball is passed around the circle of \(n\) people again. This time, the number of people missed alternates between \(a\) and \(b). How many different coloured balls of wool are now needed?
$$y=(1-\sqrt{2x})^2$$
Sum Equals Product
\(3\) and \(1.5\) are a special pair of numbers, as \(3+1.5=4.5\) and \(3\times 1.5=4.5\) so \(3+1.5=3\times 1.5\).
Given a number \(a\), can you find a number \(b\) such that \(a+b=a\times b\)?
Wool Circles
\(n\) people stand in a circle. The first person takes a ball of wool, holds the end and passes the ball to his right, missing \(a\) people. Each person who receives the wool holds it and passes the ball on to their right, missing \(a\) people. Once the ball returns to the first person, a different coloured ball of wool is given to someone who isn't holding anything and the process is repeated. This is done until everyone is holding wool.
For example, if \(n=10\) and \(a=3\):
In this example, two different coloured balls of wool are needed.
In terms of \(n\) and \(a\), how many different coloured balls of wool are needed?
Folding A4 Paper
The sides of A4 paper are in the ratio \(1:\sqrt{2}\). Let the width of the paper be 1 unit. This means that the height of the paper is \(\sqrt{2}\) units.
Therefore on the diagram, \(AE=1, BE=1, AC=\sqrt{2}\). By Pythagoras' Theorem, \(AB=\sqrt{2}\), so \(AB=AC\).
\(BF=DF=\sqrt{2}-1\) so by Pythagoras' Theorem again, \(BD=2-\sqrt{2}\). \(CD=1-(\sqrt{2}-1)=2-\sqrt{2}\). Hence, \(CD=BD\) and so the shape is a kite.
Extension
Prove that for a starting rectangle with the sides in any ratio, the resulting shape is a cyclic quadrilateral. Bézier Curves
If
$$P_0=\left(0,1\right)$$
$$P_1=\left(0,0\right)$$
$$P_2=\left(1,0\right)$$
then
$$Q_0=t\left(0,0\right)+(1-t)\left(0,1\right)=\left(0,1-t\right)$$
$$Q_1=t\left(1,0\right)+(1-t)\left(0,0\right)=\left(t,0\right)$$
and so
$$R_0=t\left(t,0\right)+(1-t)\left(0,1-t\right)=\left(t^2,(1-t)^2\right)$$
which means that
$$x=t^2$$$$y=(1-t)^2$$
or
$$y=(1-\sqrt{x})^2$$
Extension
What should \(P_0\), \(P_1\) and \(P_2\) be to get a curve with Cartesian equation
$$y=(1-\sqrt{2x})^2$$
Here's this week's collection. Answers & extensions tomorrow. Why not discuss the problems on Twitter using #SundayAfternoonMaths or on Reddit. Folding A4 Paper
A Piece of A4 paper is folded as shown: What shape is made?
Bézier Curve
1) A set of points \(P_0\), ..., \(P_n\) are chosen (in the example \(n=4\)).
2) A set of points \(Q_0\), ..., \(Q_{n-1}\) are defined by \(Q_i=t P_{i+1}+(1-t) P_i\) (shown in green).
3) A set of points \(R_0\), ..., \(R_{n-2}\) are defined by \(R_i=t Q_{i+1}+(1-t) Q_i\) (shown in blue).
.
.
.
\(n\)) After repeating the process \(n\) times, there will be one point. The Bézier curve is the path traced by this point at \(t\) varies between 0 and 1.
What is the Cartesian equation of the curve formed when:
$$P_0=\left(0,1\right)$$
$$P_1=\left(0,0\right)$$
$$P_2=\left(1,0\right)$$
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