Find the Wrong Student There are 15 student in the class and each of them has a different number 1 to 15.


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*Student #1: wrote the natural number on the board.  

*Student #2 said : This number is divisible by my number(number 2)  

*Student #3 said : This number is divisible by my number(number 3)  

*Student #4 said : This number is divisible by my number ( Number 4)  


And so on until 


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*Student #15 said : This number is divisible by my number ( Number 15 )


Student #1 is verifying the other 14 student said and he finds that all of them said it correctly except for two student with consecutive numbers. 
What is the sum of these two consecutive numbers? 
 A: The answer is $17$, as students number $8$ and $9$ are wrong.
To see this, note that if student $i$ is wrong, then student $ki$ must be wrong for every $k \ge i$.  As these will not be two consecutive numbers, this cannot be the case.  This means students $2$ through $7$ must be right.
Given $pq$, with $p$ and $q$ coprime, if student number $pq$ is wrong, then either $p$ or $q$ must be wrong as well.  (In other words, if $pq | n$, then both $p | n$ and $q | n$.)  However, these would not form two consecutive students.  This means $10$, $12$, $14$ and $15$ are right as well.
This leaves $8$ and $9$ as the only consecutive pair that could both be wrong.
A: The number must either be divisible by $4$ or by $8$ and either by $7$ or by $14$, because these are non-consecutive, hence it is divisible by $4$ and by $7$. This excludes $2,3,4,7$. In particular, it is divisible by $6$, so all of $2,3,4,5,6,7$ are divisors as well. Thus, it is also divisible by $10$, $12$ and by $15$ and so it is also divisible by $11$ and by $13$. 
This leaves only one possibility: $(8,9)$. So the sum is $17$.
