[jacks]

In a certain test there are \( n \) questions. In this test \( 2^{n-i} \) students gave wrong answers to at least \( i \) questions,

 where \( i = 1,2,3,...n. \) If total no of wrong ans given is \( 2047, \) then \( n \) is equal to

[teeteto]

Observe that the number of students that have give EXACTLY 1 wrong answer is

\( 2^{n-1}-2^{n-2}=2^{n-2} \),

which is the difference between the number of students giving at least 1 wrong answer and the number of students giving at least 2 wrong answers.

In general, the number of students that gave exactly \( i \) wrong answers for \( 1\leq i\leq n-1 \) is:

\( 2^{n-i}-2^{n-i-1}=2^{n-i-1} \),

while the number of students that gave exactly n wrong answers is \( 2^0=1 \) (because giving at least \( n \) wrong answers is the same as giving exactly \( n \) wrong answers)

Hence, adding a geometric series, we have that

\( 2047=2^{n-2}+2^{n-3}+\cdots+1=2^{n-1}-1 \)

so, \( 2^{n-1}=2048 \) and \( n=12 \).

[Luis Fuentes]

Hello

 I think that there is a small mistake in teeteto's argument.

 If the numer of studentes that gave exactly \( i \) wrong answers is:

\(  a_i=2^{ n-i-1} \) for \( i\leq 1\leq n-1,\qquad a_n=1 \)

 The total number of wrong answers is:

\(  \displaystyle\sum_{i=1}^n{}ia_i=\displaystyle\sum_{i=1}^{n-1}{}i2^{n-i-1}+n=\ldots=2^n-1 \)

 So,

\(  2047=2^n-1\quad \Rightarrow{}\quad n=11 \)

Best regards.

[teeteto]

You are right el_manco...
I counted the number of students instead of the number of mistakes... my mistake 
Thank you for correcting.

[jacks]

Thanks Admin. and Moderator