### Consecutive Numbers

An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.

### I'm Eight

Find a great variety of ways of asking questions which make 8.

### Calendar Capers

Choose any three by three square of dates on a calendar page. Circle any number on the top row, put a line through the other numbers that are in the same row and column as your circled number. Repeat this for a number of your choice from the second row. You should now have just one number left on the bottom row, circle it. Find the total for the three numbers circled. Compare this total with the number in the centre of the square. What do you find? Can you explain why this happens?

# Clocked

##### Stage: 3 Challenge Level:

Alex of The Mount School gave a clear explanation of a strategy and some thoughts on why her solutions worked. Thank you for your clear explanations Alex.

I have a few questions though ...

Is using steps of 7 hours the same as subtracting 5 hours each time and if so why?

What about mixing the number of hours between each point around the clock? Thus having steps of different lengths around the clock for example -starting at 1, step three to 4, then step four to 8 and so on. Dan of Bishops Stortford College made some suggestions and these follow Alex's work.

Firstly I decided that the hours 12 o'clock and 1 o'clock were 1 hour apart as they would be using time and not 11 hours apart as they would be numerically.

I decided to start with 1 o'clock at the top of the blank clock face. I then worked out the o'clock 3 hours after from this, which was 4 'o' clock. Then three hours after this which was 7 'o' clock I carried on until I came to 1, 4, 7, 10.
Then I found out that the sequence repeated itself (it went back to 1) so I decided to try the same strategy using 4 hours apart. My sequence was 1, 5, 9, .. then it started to repeat again.

I thought that this could be happening because both 3 and 4 divide in to 12 with no remainder, but 5 does not. So I decided to try 5:
My sequence this time was 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8 which worked.
This sequence can be changed on the clock face by starting with a new number at the top each time. Which gives us 12 different combinations.

Then I started from 1 o'clock again and insead of adding 5 hours I subtracted 5 hours. Which gave me the sequence: 1, 8, 3, 10, 5, 12, 7, 2, 9, 4, 11, 6. Any of these numbers can start at the top which gives us 12 more sequences. So I have 24 different combinations.

However I decided to go a bit farther and see what other numbers worked, 6 did not work as the sequence repeated itself however 7 did. By now I thought that a number that did not divide properly into 12 worked but then I came to 8. This did not work as it repeated so I deduced that the number must have a highest common factor of 1 and be a co-prime to 12, to be able to fit on the face correctly.

The last part of the problem asks, Can you convince us that you have them all? This is quite easily answered by probability. First though you have to get a correct answer:

Dan of Bishops Stortford College offered the following. There are some errors here but the approach is an excellent one. Perhaps some of you can help him out. How about using the discussion boards on AskNRICH?

The question says that each adjoining number has to be 3, 4 or 5 hours in difference to its neighbour. It says to arrange these around a clock face.

This can also be shown as a string of numbers: (x) y _ _ _ _ _ _ _ _ _ _ x (y) to make this work the end numbers have to differ by 3, 4 or 5 hours as well.

To find a solution that works you must use this table.

1 can be next to 4,5,6,10,9,8
2 can be next to 5,6,7,11,10,9
3 can be next to 6,7,8,12,11,10
4 can be next to 7,8,9,1,12,11
5 can be next to 8,9,10,1,2,12
6 can be next to 9,10,11,1,2,3
7 can be next to 10,11,12,2,3,4
8 can be next to 11,12,1,5,4,3
9 can be next to 12,1,2,6,5,4
10 can be next to 1,2,3,7,6,5
11 can be next to 2,3,4,8,7,6
12 can be next to 3,4,5,9,8,7

If you take the solution: (5),1,4,7,10,2,6,9,12,3,11,8,5,(1) (which is a correct solution obtained by trial and error). Then can see that the first figure (1) could have been 11 other things (2,3,4,5,6,7,8,9,10,11,12) making the first position out of 12 choices. You can then see that 4 could have been 5 other things (5,6,10,9,8) therefore making the second position out of 6 choices. This is now 12 choices with 6 possible choices after that (12x6). 7 could have been 4 other things (8,9,12,11) therefore making it out of 5 choices. This then makes it 12x6x5. If you do this for all 12 places you get 12x6x5x4x4x3x2x3x1x1x1x1=103,680. This means that there are 103,680 correct answers!!! Although this solution doesn?t answer the question: How many solutions can you find? It does say the number that there are. However I invite you to find all 103,680 solutions......