<s>[INST] A book has a few pages on which page numbers are written. Someone has torn one page out of it and now average of all page numbers is $\frac{105}{4}$ I couldn't relate this question to any of the topics specifically , I found this in a miscellaneous math problems book(non-calculus) . Here's how it goes,  A book has a few pages on which page numbers are written. Someone has torn one page out of it and now average of all page numbers is $\frac{105}{4}$. Answer the following:
(i) If the total numbers of pages in book is n then find the value of
$$\sum_{r=1}^{10} \biggl\lfloor{\frac{n+r}{r+1}}\biggr\rfloor\,.$$
 OPTIONS: 
(A)100 (B)107 (C)105 (D)82
(ii) If the line $x+y=\bigl\lfloor\frac{n}{3}\bigr\rfloor$ is drawn ,then the total number of points with integral co-ordinates enclosed within the region bounded by $x=0,y=0$ and $x+y=\bigl\lfloor\frac{n}{3}\bigr\rfloor$ is -----?
(A)105 (B)153 (C)59 (D)78 

STATUS: No clue how to start.
Any help is welcome.
 [/INST] More on how to deduce the value of $n$. The average page number before a page is torn is
$$ \frac{1+2+\cdots+n}{n}=\frac{n+1}{2}$$
Since two consecutive page numbers are removed, let the two page numbers be $2k-1$ and $2k$, where $2\le 2k\le n$, then
$$\begin{align}
n\cdot\frac{n+1}{2}-(n-2)\frac{105}{4} =& 2k-1+2k\\
\frac{n}{2}(n+1)-\left(\frac{n}{2}-1\right)\left(52+\frac{1}{2}\right) =& 4k-1
\end{align}$$
Since $\frac{n}{2}$ and $k$ are integers, to make the above left hand side an integer, we must have $\frac{n}{2}-1$ even, or $\frac{n}{2}$ is an odd number. Let $n=4m+2$, where $m$ is an integer.
$$\begin{align}
(2m+1)(4m+2+1)-105m+1 =& 4k
\end{align}$$
Again, since $m$ and $k$ are integers, to make the above left hand side an even number, $m$ must be an even number. Let $m=2p$, where $p$ is an integer. Then $n=8p+2$.
$$\begin{align}
(4p+1)(8p+3)-210p+1 =& 4k\\
32p^2-190p+4 =& 4k
\end{align}$$
Again, since $p$ and $k$ are integers, to make the above left hand side a multiple of 4, $p$ must be an even number. Let $p=2q$, where $q$ is an integer. Then $n=16q+2$.
$$\begin{align}
128q^2-380q+4 =& 4k\\
32q^2-95q+1 =& k
\end{align}$$
Considering the conditions for the value of $k$,
$$\begin{array}{rcl}
2 \le& 2k &\le n\\
2 \le& 2(32q^2-95q+1) &\le 16q+2\\
0 \le& 2q(32q- 95) &\le 16q\\
0 \le& 32q- 95 &\le 8\\
95 \le& 32q &\le 103\\
3-\frac{1}{32} \le& q &\le 3+\frac{7}{32}\\
\end{array}$$
Therefore, we only have
$$\begin{align}
q =& 3\\
n =& 16q+2 = 50\\
2k=& 8\\
2k-1 =& 7
\end{align}$$
</s>