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Thread: [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 $\displaystyle n$ questions. In the test, $\displaystyle 4^{n - k}$ students gave wrong answers to at least $\displaystyle k$ questions, where $\displaystyle k = 1,2,\mbox{...},n$. 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 $\displaystyle k$ questions wrong is $\displaystyle 4^{n-k}$. Hence the number that actually get exactly $\displaystyle k$ wrong, $\displaystyle N_k$ , must be the number that get at least $\displaystyle k$ wrong minus those that get at least $\displaystyle (k+1)$ wrong. So we have

    $\displaystyle N_k = 4^{n-k} - 4^{n-k-1}$ .

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

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

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

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

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

    $\displaystyle = 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 $\displaystyle m= k+1$),

    $\displaystyle = 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)} $,

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

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

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

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

    (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

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

    and solving gives us that

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


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