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 $\frac{105}{4}$

Question

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.

Answer 1

OK, so here is a solution to (i) in case we know that when a page is torn out and odd and the following even page number is removed.

Given the book has $n$ pages the sum of the page numbers will be $$ T_n=1+2+...+n=\frac{n(n+1)}{2} $$ known as the $n$'th Triangular Number. So assume that page numbers $x$ and $x+1$ have been torn out where $x$ is odd. Then we can write $$ \frac{105}{4}=\tfrac{1}{n-2}\left(T_n-(2x+1)\right) $$ Plugging in the formula for $T_n$ from above and solving for $x$ then yields $$ x=\frac{1}{4}n^2-\frac{103}{8}n+\frac{103}{4} $$ which is a quadratic expression in $n$ with zeros $n=\frac{103\pm\sqrt{8961}}{4}$ which approximately is $n=2.08$ and $n=49.42$. On the other hand it is obvious that $x<n$ which then in turn yields a quadratic inequality in $n$ that can be solved to see that $n<\frac{111+\sqrt{10673}}{4}\approx 53.58$.

Having a slightly closer look at the expression for $x$ one realizes that $n-2$ must be divisible by $8$ for $x$ to be an integer. So unless we take $n=2$ (which actually works) we must have $49.42<n=50<53.58$ for $n$ to be an integer satisfying all requirements in that interval. Plugging $n=50$ into the expression for $x$ then yields $x=7,x+1=8$, and you can check as I did in the comments that the average of the remaining pages is really $\frac{105}{4}$. With $n=50$ one gets $$ \sum_{i=1}^{10}\left\lfloor\frac{50+r}{r+1}\right\rfloor=105 $$ So option (C) answers the first question correctly.

Part (ii)

Still using $n=50$ we get $\lfloor\frac{50}{3}\rfloor=16$ so that the line is $x+y=16$ or $y=16-x$. Together with the axes $y=0$ and $x=0$ this encloses a triangle with $17$ lattice points on the line $y=16-x$ since you can start from $(0,16)$ and move down right step-by-step to 'visit' 17 lattice points before you hit the x-axis. Similarly you hit $16$ lattice points following the same procedure from $(0,15)$. With this we get the answer to (ii) which is: $$ T_{17}=17+16+...+1=\frac{17\cdot 18}{2}=153 $$ So option (B) answers the second question correctly.

Answer 2

For 1, we have to assume the pages are consecutively numbered from $1$ to $n$ before a single page is removed (which removes two page numbers). The average of all the page numbers is $\frac {n+1}2$. The remaining number of pages must be a multiple of $4$ so we can get the denominator in the average. Clearly it will not move the average much, so the average is about $26$ and $n$ is about twice that. So we guess $n=54$ will work. The sum of all the pages is $\frac {54 \cdot 55}2=1485$. The sum of the remaining pages if $52 \cdot \frac {105}4=1365$, so the sum of the removed page numbers is $120$. Unfortunately there is no page with that sum. Trying $n=50$ gives a starting sum of $1275$ and an ending sum of $1260$, leading to removal of the page that includes numbers $12$ and $13$. It is then arithmetic to evaluate the sum, getting $105$

An approach without guessing, but with more work is as follows: If we let $m$ be the page number removed, we have $$\frac {\frac {n(n+1)}2-m}{n-1}=\frac {105}4\\ \frac {n(n+1)}2-m=\frac {105(n-1)}4\\2n^2+2n-4m=105n-105\\2n^2-103n+105-4m=0\\n=\frac 14\left(103 +\sqrt{103^2+32m-620}\right)$$ and searching for $m$ that will make the square root an integer will get there.

Answer 3

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

$$\begin{align} q =& 3\\ n =& 16q+2 = 50\\ 2k=& 8\\ 2k-1 =& 7 \end{align}$$