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    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?
    • Study Helper
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    Yes, it is possible!
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    (Original post by Chlorophile)
    Yes, it is possible!
    Ok do it o_O
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    (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?
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    (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!

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

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    If you start on day 1, then you must repeat a pattern on day 378.
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    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.
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    (Original post by RVNmax)
    ....
    ..nevermind, looks like you noticed!
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    (Original post by FireGarden)
    ..nevermind, looks like you noticed!
    Haha thanks. I thought I realised before anyone else would have but I guess not...
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    Who does this? It's either 1 or 2 not both combined randomly.
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    (Original post by RVNmax)
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    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  T_n 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  T_n = T_{n-1} + T_{n-2} with T_1 = 1 and T_2 = 2.

    It's not hard to see that  T_n = F_{n+1} where  F_n is the nth fibonacci number. Hence  T_{13} = F_{14} = 377

    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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