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# Maths prob 2 Watch

1. When I go up my stairs in a hurry, I take some steps two-at-a-time.

This morning, I climbed my thirteen steps in this pattern: 1, 2, 1, 2, 1, 1, 2, 1, 2.
Yesterday the pattern was 1, 1, 1, 1, 1, 1, 2, 2, 1, 2.

How many days do you think I can go before I have to repeat a pattern?

Easiest way to do this?
2. Yes, it is possible!
3. (Original post by Chlorophile)
Yes, it is possible!
Ok do it o_O
4. (Original post by Zenarthra)
Ok do it o_O
Wait, are you asking for help or are you posing this as a question to the TSR community?
5. (Original post by Chlorophile)
Wait, are you asking for help or are you posing this as a question to the TSR community?
It was both actually ^^
Wanted to know if it was possible and solutions/ideas conveyed by the community.
SO WE CAN HELP EACH OTHER SOLVE IT!! YEAHH!

Spoiler:
Show
Dat waz ghey
6. Partition around either climbing one step first or two steps first.. You'll end up with a recurrence. And it should look familiar.
7. Actually i shouldn't asked if it was possible, didnt even think...
8. Nice little puzzle. Answer (note, not solution..) below, if you'd like.

Spoiler:
Show
If you start on day 1, then you must repeat a pattern on day 378.
9. Min. number of steps needed is 7 unless you can go more than two-at-a-time. There would be 7 ways of doing it in 7 steps. [2+2+2+2+2+2+1 or 2+2+2+2+2+1+2 etc.]
Max is 13 steps. - 1 way!

Each of the ones in the middle can be done as well:
8 steps can be done with 5 doubles and 3 singles. treat as 8 slots with a combination of 3 of one item and 5 of another, so 8C3 = 56 (= 8C5)
try to finish before reading solution
9 steps can be done with 4 doubles and 5 singles. so 9C4 = 126 ways of doing it 9 steps.
10 steps in 10C3 =120 ways
11 in 11C2 = 55 ways
12 in 12 ways.

add altogether, so 7 + 1 + 56 + 126 +120 + 55 + 12 = 377 ways by my calculations. 377 days without repeating pattern. repeat on 378th.
10. (Original post by RVNmax)
....
..nevermind, looks like you noticed!
11. (Original post by FireGarden)
..nevermind, looks like you noticed!
Haha thanks. I thought I realised before anyone else would have but I guess not...
12. Who does this? It's either 1 or 2 not both combined randomly.
13. (Original post by RVNmax)
Spoiler:
Show
Min. number of steps needed is 7 unless you can go more than two-at-a-time. There would be 7 ways of doing it in 7 steps. [2+2+2+2+2+2+1 or 2+2+2+2+2+1+2 etc.]
Max is 13 steps. - 1 way!

Each of the ones in the middle can be done as well:
8 steps can be done with 5 doubles and 3 singles. treat as 8 slots with a combination of 3 of one item and 5 of another, so 8C3 = 56 (= 8C5)
try to finish before reading solution
9 steps can be done with 4 doubles and 5 singles. so 9C4 = 126 ways of doing it 9 steps.
10 steps in 10C3 =120 ways
11 in 11C2 = 55 ways
12 in 12 ways.

add altogether, so 7 + 1 + 56 + 126 +120 + 55 + 12 = 378 ways by my calculations.
I spent ages trying to find where you went wrong, and you added the bloody numbers wrong at the end! It's also not a very nice solution!

Let be the number of ways we can walk up n steps. Partitioning around whether we climb one step or two steps first, we get the recurrence with and .

It's not hard to see that where is the nth fibonacci number. Hence

And so, starting on day 1, day 378 will be the day that contains the repeat (as already said, but I thought I'd provide a solution since another has already been provided).

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Updated: July 12, 2014
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