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In a certain test there are n questions, in this test $2^{n−i}$ students gave wrong answer to at least i questions where i=1,2,3,…,n. If the total number of wrong answers given is 2047,what is the value of n?

My attempt:From the question $2^{n−1}$ students gave wrong answer to at least one question,$2^{n-2}$ students to at least 2 and so on.Hence there's one student who answered all n questions wrong.Now $2^{n-1}-2^{n-2}$ gives number of wrong answers by those students who got one question wrong.Similarly $2^{n-2}-2^{n-3}$ gives the number of wrong answers by students who got 2 questions wrong.On adding all such cases I arrived at $2^{n-1}-1=2047$ which gave me n=12 .Where did i go wrong?

GrandAlpha
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1 Answers1

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$2^{n-n}=1$ student got all answers wrong. So he contributed $1\times n=n$ wrong answers.

Now, given at least $2^{n-i}$ students answered 'i' questions wrong. So the number of students who gave 'i' answers (NOT at least i) wrong is $2^{n-i}-2^{n-i-1}$.

Therefore, total number of wrong answers contributed by $2^{n-i}-2^{n-i-1}$ students who answered i questions wrong is $(2^{n-i}-2^{n-i-1})i$

Now, sum of wrong answers $\sum\limits_{i=1}^{n-1}{(2^{n-i}-2^{n-i-1})i} + n =2047$

Simplifying LHS, $\sum\limits_{i=1}^{n-1}{(2^{n-i}-2^{n-i-1})i} + n = \sum\limits_{i=1}^{n-1}{2^{n-i-1}i} + n = 2^{n-1}\sum\limits_{i=1}^{n-1}{2^{-i}i} + n$

$\sum\limits_{i=1}^{n-1}{2^{-i}i}$ is nothing but arithmetico-geometrico progression with sum $2^{1-n}(-n+2^n-1)$ (Verify with induction)

Then $2^{n-1}\sum\limits_{i=1}^{n-1}{2^{-i}i} + n = 2^{n-1}\times2^{1-n}(-n+2^n-1) + n = 2^n-1 = 2047$

That is, $2^n=2048$

Therefore, n =11.

Jesse P Francis
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    Nice solution. I had to do an induction proof to verify the statement $$P(n): \sum_{k = 1}^{n - 1} \frac{k}{2^k} = \frac{2^n - n - 1}{2^{n - 1}}$$ for $n \geq 2$. Once I did that, everything became clear. – N. F. Taussig Apr 10 '15 at 23:22