Puzzles
Shooting hoops
You spend an afternoon practising throwing a basketball through a hoop.
One hour into the afternoon, you have scored less than 75% of your shots. At the end of the afternoon, you have score more than 75% of your shots.
Is there a point in the afternoon when you had scored exactly 75% of your shots?
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If there is not a point when you had scored exactly 75%, then there must be a shot which you score which took your average over 75%. In other words, there must be an $a$ and a $b$ so that:
$$\frac{a}{b}<\frac{3}{4}<\frac{a+1}{b+1}$$
Rearranging these inequalities gives:
$$4a<3b$$
$$3b+3<4a+4$$
Taking 3 from the bottom inequality gives:
$$4a<3b$$
$$3b<4a+1$$
Combining the inequalities gives:
$$4a<3b<4a+1$$
Therefore \(3b\) must be an integer which is between \(4a\) and \(4a+1\). But \(4a\) and \(4a+1\) are consecutive numbers so there is no other integer between them.
This means that this situation is impossible and there must have been a time when you had scored exactly 75% of your shots.
Extension
You spend another afternoon practising throwing a basketball through a hoop.
One hour into the afternoon, you have scored more than 75% of your shots. At the end of the afternoon, you have score less than 75% of your shots.
Is there a point in the afternoon when you had scored exactly 75% of your shots?
Rugby scores
In a rugby (union) match, 3 point are scored for a kick, 5 for a try and 7 for a converted try. This scoring system means that some total scores can be achieved in different combinations, while others can be achieved in only one way.
For example, 14 can be scored in two ways (three kicks and a try; or two converted tries), while 8 can only be achieved in one way (try and a kick).
What is the highest score which can only be made in one way?
What is the highest score which can be made in two ways?
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11 is the highest score which can only be made in one way (try and two kicks).
Each of 12, 13 and 14 can be made in two ways:
- 12 - try and converted try; or four kicks
- 13 - converted try and two kicks; or two tries and a kick
- 14 - three kicks and a try; or two converted tries
Now any score higher than 14 can be made by adding kicks to one of these three. This gives at least two ways of making every score higher than this.
By a similar argument, it can be shown that 16 is the highest score which can only be made in two ways.
For the highest score which can be made in \(n\) ways, see the OEIS sequence
A261155.
Extension
Which is the highest rugby score such that if the number of tries scored is known, you can always tell how many of these tries were converted?
Integer part
Let \(\lfloor x\rfloor \) denote the integer part of \(x\) (eg. \(\lfloor 7.8\rfloor =7\)).
When are the following true:
a) \(\lfloor x+1\rfloor = \lfloor x\rfloor + 1\)
b) \(\lfloor nx\rfloor = n\lfloor x\rfloor\) (where \(n\) is an integer)
c) \(\lfloor x+y\rfloor = \lfloor x\rfloor +\lfloor y\rfloor \)
d) \(\lfloor xy\rfloor = \lfloor x\rfloor \lfloor y\rfloor \)
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a) Always
b) When \(n<\frac{1}{f_x}\), where \(f_x\) is the fractional part of \(x\).
c) When the fractional parts of \(x\) and \(y\) add up to less than one.
d) Let \(f_x\) and \(f_y\) be the fractional parts of \(x\) and \(y\) (respectively).
$$\lfloor xy\rfloor = \lfloor (\lfloor x\rfloor +f_x)(\lfloor y\rfloor +f_x)\rfloor $$
$$=\lfloor \lfloor x\rfloor \lfloor y\rfloor +f_x\lfloor y\rfloor +f_y\lfloor x\rfloor +f_yf_x\rfloor $$
$$=\lfloor x\rfloor \lfloor y\rfloor +\lfloor f_x\lfloor y\rfloor +f_y\lfloor x\rfloor +f_yf_x\rfloor $$
This will be equal to \(\lfloor x\rfloor \lfloor y\rfloor \) when \(\lfloor f_x\lfloor y\rfloor +f_y\lfloor x\rfloor +f_yf_x\rfloor =0\).
For this to be true, it is necessary (but not sufficient) that \(f_y<\frac{1}{x}\) and \(f_x<\frac{1}{y}\).
Extension
Show that
$$\lfloor x\rfloor +\left\lfloor x+\frac{1}{n}\right\rfloor +\left\lfloor x+\frac{2}{n}\right\rfloor +...+\left\lfloor x+\frac{n-1}{n}\right\rfloor =\lfloor nx\rfloor.$$
Make the sums
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums reading across and down are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, \(4+3\times2\) is 14, not 10.
| + | | - | | = 4 |
| + | | - | | × | |
| - | | × | | = 27 |
| - | | × | | ÷ | |
| × | | ÷ | | = 16 |
=
2 | | =
8 | | =
6 | |
Show answer & extension
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| 1 | + | 5 | - | 2 | = 4 |
| + | | - | | × | |
| 7 | - | 4 | × | 9 | = 27 |
| - | | × | | × | |
| 6 | × | 8 | - | 3 | = 16 |
=
2 | | =
8 | | =
6 | |
Extension
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums reading across and down are correct.
| + | | - | | = 5 |
| - | | - | | - | |
| + | | ÷ | | = 5 |
| + | | ÷ | | × | |
| + | | × | | = 99 |
=
0 | | =
1 | | =
18 | |
Rotating round table
At a large dinner, 24 people are to sit evenly spaced around a round table. Place cards are laid to show where everyone should sit. Unfortunately nobody notices the name cards and the guests sit down with nobody in the correct seat.
Show that it is possible to rotate the table so that at least two people will be in the correct seats.
Show answer & extension
Hide answer & extension
Let \(p(n)\) denote the number of people in the correct seat when the table is rotated by \(n\) people. From the question, we know that:
$$p(0)=0$$
As every guest can be put into the correct place by one rotation, we know that:
$$p(0)+p(1)+...+p(23)=24$$
As \(p(0)=0\):
$$p(1)+...+p(23)=24$$
This sum has 23 positive integers adding up to 24, so one of the rotations must lead to at least two people being in the correct places.
Extension
If the 24 guests sat randomly and one person was in the correct seat, could you still rotate the table so that two people are correctly seated?
121
Find a number base other than 10 in which 121 is a perfect square.
Fill in the digits
Can you place the digits 1 to 9 in the boxes so that the three digit numbers formed in the top, middle and bottom rows are multiples of 17, 25 and 9 (respectively); and the three digit numbers in the left, middle and right columns are multiples of 11, 16 and 12 (respectively)?
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?
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