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Math Help - [SOLVED] Find number of questions?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Find number of questions?

    In a certain test there are n questions. In the test, 4^{n - k} students gave wrong answers to at least k questions, where k = 1,2,\mbox{...},n. If the total number of wrong answers is 341, then find the value of n.
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  2. #2
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    Solution

    Here's how to do it:

    Spoiler:


    The question states that the number of students with at least k questions wrong is 4^{n-k}. Hence the number that actually get exactly k wrong, N_k , must be the number that get at least k wrong minus those that get at least (k+1) wrong. So we have

    N_k = 4^{n-k} - 4^{n-k-1} .

    So if we let N_q be the total number of wrongly answered questions then

    N_q = n + \sum_{k=1}^{n-1} k N_k ,

    = n + \sum_{k=1}^{n-1} k (4^{n-k} - 4^{n-k-1}),

    = n + \sum_{k=1}^{n-1} k 4^{n-k} - \sum_{k=1}^{n-1} k 4^{n-k-1},

    = n + \sum_{k=1}^{n-1} k 4^{n-k} - \sum_{k=1}^{n-1} (k+1) 4^{n-(k+1)} + <br />
\sum_{k=1}^{n-1} 4^{n-(k+1)},

    = n + \sum_{k=1}^{n-1} k 4^{n-k} - \sum_{m=2}^{n} m 4^{n-m} + \sum_{k=1}^{n-1} 4^{n-(k+1)} (letting m= k+1),

    = n + \sum_{k=1}^{n-1} k 4^{n-k} - \sum_{m=2}^{n} m 4^{n-m} + \sum_{k=1}^{n-1} 4^{n-(k+1)} ,

    = n + 4^{n-1} -  n + \sum_{k=1}^{n-1} 4^{n-(k+1)} ,

    = \sum_{k=0}^{n-1} 4^{n-(k+1)} ,

    = \sum_{p=0}^{n-1} 4^{p} (letting p = n-(k+1)),

    = \frac{4^{n}-1}{4-1} = \frac{1}{3}(4^n -1).

    (I know I could have explicitly written out the sum, 'spot the pattern' - as I did on paper, but decided this was typographically easier than using an array).

    So we have that

    341 = \frac{1}{3}(4^n -1)

    and solving gives us that

    \color[rgb]{0,0,1} \boxed{n = 5}.


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