Puzzles
21 December
There are ten ways to make a list of four As and Bs that don't contain an even* number of consecutive As:
- B,B,B,B
- A,B,B,B
- B,A,B,B
- B,B,A,B
- B,B,B,A
|
- A,B,A,B
- A,B,B,A
- B,A,B,A
- A,A,A,B
- B,A,A,A
|
How many ways are there to make a list of eleven As and Bs that don't contain an even number of consecutive As?
* We don't count 0 consecutive As as being an even number of consecutive As.
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Valid sequences of length \(n\) can be made by doing one of the following:
- Taking a valid sequence of length \(n-1\) and adding a B to the end of it.
- Taking a valid sequence of length \(n-1\) that ends in B and adding a A to the end of it.
- Taking a valid sequence of length \(n-2\) that ends in A and adding two As to the end of it.
Therefore,
the number of sequences of length \(n\) that end in B is equal to the total number of sequences of length \(n-1\);
and the number of sequences of length \(n\) that end in A is equal to the number of sequences of length \(n-1\) that end in B plus the number of sequences of length \(n-2\) that end in A.
Using this we see that:
| \(n\) | number of sequences that end A | number of sequences that end B |
| 3 | 3 (A,A,A and A,B,A and B,B,A) | 3 (B,B,B and A,B,B and B,A,B) |
| 4 | 4 | 6 |
| 5 | 9 | 10 |
| 6 | 14 | 19 |
| 7 | 28 | 33 |
| 8 | 47 | 61 |
| 9 | 89 | 108 |
| 10 | 155 | 197 |
| 11 | 286 | 352 |
The total number of valid sequences of length 11 is 638.
2 December
Holly adds up the first six even numbers, then adds on half of the next even number. Her total is 49.
Next, Holly adds up the first \(n\) even numbers then adds on half of the next even number. This time, her total is 465124. What is \(n\)?
Show answer & extension
Hide answer & extension
If we add up the first \(n\) even numbers then add on half of the next even number, we get \((n+1)^2\). This means that Holly added up the first \(\sqrt{465124}-1\) or 681 even numbers.
Extension
Can you show why adding up the first \(n\) even numbers and half of the next even number gives \((n+1)^2\)?
15 December
There are 3 even numbers between 3 and 9.
What is the only odd number \(n\) such that there are \(n\) even numbers
between \(n\) and 729?
Show answer & extension
Hide answer & extension
There are \((729-n)/2\) even numbers between \(n\) and 729, and so we want to solve \((729-n)/2=n\). The solution of this is 243.
Extension
For which odd numbers \(N\) does there exist an odd number \(n\) such that there are \(n\) even numbers between \(n\) and \(N\)?
