Post Number:#3 by ahmedittihad » Mon Nov 27, 2017 4:41 pm
You're actually misinterpreting the question.
The problem basically gives us that $(x+1) + (x+2) + ... + (x+y) =976 $. With $x+1$ being the first missing page and $x+y$ the last missing page. So we need to find $y$.
$(x+1) + (x+2) + ... + (x+y) = xy + \dfrac {y(y+1)}{2}=976$
So, $y(x+ \dfrac {y+1}{2})=976$.
Now, $\dfrac {y+1}{2}$ is an integer. So $y$ must be odd. Let's look at the divisors of $976$. They are $1,2,4,8,16,61,122,244,488,976$. The question mentioned several pages so $y \neq 1$ and $y$ is odd and the only odd number except $1$ in the set is $61$. So we get, $y=61$
And there are $61$ pages.
Frankly, my dear, I don't give a damn.