Puzzles
More doubling cribbage
Brendan and Adam are playing lots more games of
high stakes cribbage: whoever
loses each game must double the other players money. For example, if Brendan has £3 and Adam has £4 then Brendan wins, they will have £6
and £1 respectively.
In each game, the player who has the least money wins.
Brendan and Adam notice that for some amounts of
starting money, the games end with one player having all the money; but for other amounts, the games continue forever.
For which
amounts of starting money will the games end with one player having all the money?
Show answer & extension
Hide answer & extension
If Adam has £\(a\) and Brendan has £\(b\), we will write this as £\(a\):£\(b\).
First, we can take
\(a\) and \(b\) to have no common factors, as dividing both by a common factor gives an equivalent starting point. For example,
£3:£6 and £6:£12 will have exactly the same behaviour (imagine £6 and 3 £2 coins).
£\(a\):£\(b\) is also clearly equivalent to £\(b\):£\(a\) (but with the two players swapping places).
A game starting £\(a\):£\(b\) will end with one player having the money if \(a+b\) is a power of two. This is because:
If \(a+b=2^n\), then if \(n=1\), either the game has already ended or it sits at £1:£1 and is about to end. If \(n>2\), then the starting
position can be written as £\(2^n-k\):£\(k\) with \(k<2^n-k\). After another game this will be £\(2^n-2k\):£\(2k\). This is equivalent to
£\(2^{n-1}-k\):£\(k\). Therefore by
induction, if \(a+b=2^n\) then the
game ends with one player having all the money.
It can also be shown by induction, that if a game ends then it must be £\(a\):£\(b\)
with \(a+b=2^n\).
Extension
What would happen if the losing player has to triple the winning player's money?
Doubling cribbage
Brendan and Adam are playing high stakes cribbage: whoever loses each game must double the other players money. For example, if Brendan has £3 and Adam has £4 then Brendan wins, they will have £6 and £1 respectively.
Adam wins the first game then loses the second game. They then notice that they each have £180. How much did each player start with?
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Hide answer & extension
Working backwards: before the second game, Brendan must have had £90 and so Adam had £270.
Before the first game, Adam must have had £135, so Brendan had £225.
Extension
After the next game, one player will have all the money and no more games can be played. Hence £135 and £225 lead to a finite number of games being played.
If the player with the most money always loses, which starting values £\(A\) and £\(B\) will lead to finite and infinite numbers of games?
The taxman
In a very strange country, the tax system works as follows.
£1, £2, £3 up to £12 are available.
You pick an amount. You keep this amount, but the taxman takes any factors of it. You cannot pick any amount without a factor.
This continues until you can take no more money. The taxman gets any remaining money.
For example, you might play as follows:
- Take £12. Taxman gets £1, £2, £3, £4, £6.
- Take £10. Taxman gets £5.
- You cannot take anything. Taxman gets £7, £8, £9, £11.
In this example, you end with £22 and the taxman ends with
£56.
Is it possible to get more money than the taxman? What is the most you can get?
Show answer & extension
Hide answer & extension
The maximum you can get is £50, with the taxman getting
£28. Here is
how to get it:
- Take £11. Taxman gets £1.
- Take £9. Taxman gets £3.
- Take £8. Taxman gets £2, £4.
- Take £10. Taxman gets £5.
- Take £12. Taxman gets £6.
- You cannot take anything. Taxman gets £7.
Extension
Can you always beat the taxman if £1 up to £\(n\) are
available?
No change
"Give me change for a dollar, please," said the customer.
"I'm sorry," said the cashier, "but I can't do it with the coins I have. In fact, I can't change a half dollar, quarter, dime or nickel."
"Do you have any coins at all?" asked the customer.
"Oh yes," said the cashier, "I have $1.15 in coins."
Which coins are in the cash register?
(The available coins are 50¢, 25¢, 10¢ 5¢ and 1¢.)
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¢?
Show answer & extension
Hide answer & extension
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?
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.
Show answer & extension
Hide answer & extension
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?
Ninety nine
In a 'ninety nine' shop, all items cost a number of pounds and 99 pence. Susanna spent £65.76. How many items did she buy?
Show answer & extension
Hide answer & extension
Every item bought will cause the pence in the total cost to fall by 1. So to spend £65.76, Susanna must have bought 24 items.
Extension
What is the smallest amount Susanna could spend for which we could not tell how many items she bought?
