A problem with the page numbers

Relevant wiki: Sum of n, n², or n³

The sum of the $n$ pages is $\frac{n(n+1)}{2}$ (see this wiki ). Let $k$ be the number on the front of the page torn out. So the sum of the numbers of the page torn out is $k+k+1=2k+1$ (note that this number is always odd). Then the sum of the page numbers of the remaining pages is

$525=\frac{n(n+1)}{2}-(2k+1)$

This looks complicated now, but we can estimate. Instead of the big expression on the right, use:

$\begin{aligned} 525 &\approx \frac{n^2}{2} \\ 1050 &\approx n^2 \\ n &\approx 32 \end{aligned}$

Suppose the number of pages is $n=32.$ Then the sum of the page numbers is $\frac{(32)(33)}{2}=528$

This means that the sum of the torn out page numbers is $1+2=\boxed{3}.$ This could be possible if the first page is torn out (page numbers 1 and 2).

Is it possible that there is more than one solution?

First, note that there must be an even number of page numbers. Suppose that there were $n=30$ page numbers. Then the sum of the page numbers is $\frac{(30)(31)}{2}=465.$ This is too low, so it wouldn't be possible.

Now suppose that there were $n=34$ page numbers. Then the sum of the page numbers is $\frac{(34)(35)}{2}=595.$ This would make the sum of the torn out page numbers $595-525=70.$ However, the sum of the torn out page numbers must be odd.

Now suppose that there were $n=36$ page numbers. Then the sum of the page numbers is $\frac{(35)(36)}{2}=666.$ This would make the sum of the torn out page numbers $666-525=141.$ This is odd, but the page numbers would have to be 70 and 71. Not only is this not possible because of the numbering scheme (lower number is always odd), but these numbers are higher than the highest page number! We can then see that any number of higher pages is also impossible.

Now, if you're really up for a challenge, see if it is possible that there is ever more than one solution with $n$ page numbers, one page torn out, and the sum of remaining page numbers $S.$

This solution derives the specific Diophantine Equations, and is a more advanced answer for the curious.

First, as in Andy's solution we assume the book has $n$ numbered page sides and we tear out the page starting with $k,$ and we want to solve:

$525 = \frac{n(n+1)}{2} - (2k+1)$

which after a little rearrangement becomes

$4k = n^2 + n - 1052$

Note since $n$ and $k$ are both integers, and 1052 is divisible by 4, that $n^2+n$ must be divisible by 4. Now, consider every possible value of $n$ from 0 to 3 modulo 4:

$\begin{aligned} 0^2 + 0 = 0 &\equiv 0 \bmod 4 \\ 1^2 + 1 = 2 &\equiv 2 \bmod 4 \\ 2^2 + 2 = 6 &\equiv 2 \bmod 4 \\ 3^2 + 3 = 12 &\equiv 0 \bmod 4 \end{aligned}$

This means that, for all integers $t,$ $n$ must equal either $4t$ (when $n$ is 0 modulo 4) or $4t+3$ (when $n$ is 3 modulo 4).

First consider $n = 4t :$

$\begin{aligned} 4k &= n^2 + n - 1052 \\ 4k &= (4t)^2 + (4t) - 1052 \\ 4k &= 16t^2 + 4t - 1052 \\ k &= 4t^2 + t - 263 \end{aligned}$

Now, when $n = 4t + 3 :$

$\begin{aligned} 4k &= n^2 + n - 1052 \\ 4k &= (4t+3)^2 + (4t+3) - 1052 \\ 4k &= 16t^2 + 24t + 9 + 4t + 3 - 1052 \\ 4k &= 16t^2 + 28t - 1040 \\ k &= 4t^2 + 7t - 260 \end{aligned}$

Note that $k$ (the page number torn) must be positive, which holds in both equations only when $t \geq 8 .$

When $t = 8 ,$ using the first equation yields $n = 4t = 32$ and $k = 4(64) + 8 - 263 = 1 .$ This corresponds to tearing the page starting with 1 out of a book numbered to 32.

When $t = 8 ,$ using the second equation yields $n = 4t + 3 = 35$ and $k = 4(64) + 7(8) - 260 = 52.$ This corresponds to tearing page 52 out of a book numbered to 35, which is impossible.

$t = 9$ gets impossible values in both cases. Note, since $t^2$ will outgrow $t,$ that the page number being torn will be larger than the number of pages in the book for all $t > 8 .$ Therefore tearing page 1 out of a book numbered to 32 is the only solution.

Let $N$ be the number of sheets in the magazine (numbered 1 through $2N$ ), then the sum of all page numbers is $\Sigma = \tfrac12 (2N) (2N+1) = 2N^2 + N.$ Let $1 \leq p \leq N$ be the number of the missing sheet; the sum of its page numbers is $(2p-1) + 2p = 4p-1$ . Thus we must solve $2N^2 + N - (4p-1) = S,$ where $S$ is the sum of remaining page numbers. For the minimum value $p = 1$ and the maximum value $p = N$ we find solutions $2N_-^2 + N_- - (S + 3) = 0\ \ \ \ \therefore\ \ \ \ N_- = \tfrac14(\sqrt{25 + 8S} - 1).$ $2N_+^2 - 3N_+ - (S + 1) = 0\ \ \ \ \therefore\ \ \ \ N_+ = \tfrac14(\sqrt{17 + 8S} + 3).$ The difference between these extremes is $N_+ - N_- = \tfrac14(4 - (\sqrt{25 + 8S} - \sqrt{17 + 8S})) < 1,$ so that there is at most one integer value between these two extremes; this integer provides the solution, if any exists.

With the given value $S = 525$ , we have $N_- = \tfrac14(\sqrt{4225}-1) = 16;$ there is no need to check that $N_+ \approx 16.98$ because we already have a solution in $N = N_- = 16$ , and we know that $N_+ < 16 + 1 = 17$ .

Thus there are 16 sheets (32 pages), and the missing sheet is the first one, $p = 1$ , with sum of page numbers $1 + 2 = \boxed{3}$ .

Interesting observation

The sum of remaining page numbers is a function $S(N,p)$ for $1 \leq p \leq N$ which can be defined recursively:

• A magazine with no sheets is a magazine with one sheet, which is ripped out. Thus $S(1,1) = 0$ .

• Put a ripped out page back in the magazine, and instead rip out the one before that. Since the two ripped-out page numbers are now two less than before, this increases the sum by 4. Thus $S(N,p-1) = S(N,p) + 4$ .

• If the first sheet (pages 1 and 2) are ripped out, put them back in; this increases the sum by 3. Now add a page to the end of the magazine (increasing $N$ by 1), and rip it out. We find $S(N+1,N+1) = S(N,1) + 3$ .

This pattern is illustrated below:

$\begin{array}{rr} (N,p) & S(N,p) \\ \hline (1,1) & 0 \\ \hline (2,2) & 3 \\ (2,1) & 7 \\ \hline (3,3) & 10 \\ (3,2) & 14 \\ (3,1) & 18 \\ \hline (4,4) & 21 \\ (4,3) & 25 \\ \vdots \end{array}$

From the assumptions, since there are numbers on both sides of every page in this magazine, the final page number of this magazine is an even number. And since a page has been torn from the magazine, the sum of all the page numbers including that page is greater than 525. If $n$ is the final page number of this magazine:

$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}>525$ $n^{2}+n-1050>0$ $Relevant \space answer:\quad n>\frac{-1+\sqrt{4201}}{2}\approx 31.91$

Edit: The remainder of my explanation mirrored Andy Hayes's a little too much, so I've thought of a different one.

The sum of the page numbers on the missing page is the sum of the pages numbers of the intact magazine minus 525. This has a maximum, the sum of the page numbers on the last page. So, using the same $n$ :

$\frac{n(n+1)}{2}-525\leq n+(n-1)$ $n^{2}-3n-1048\leq 0$ $Relevant \space answer:\quad n\leq \frac{3+\sqrt{4201}}{2}\approx 33.91$

Remember that n is an even integer. And the only even integer that fits those two parameters is $32$ . Therefore:

$\frac{32*33}{2}-525=\color{#3D99F6}{3}\color{#333333}\quad\Box$

One can imply from this that the first page was removed (1+2).

the sum of consecutive numbers from 1 that is closest to (but greater than) the sum of the remaining pages (525) is 528. this gives a difference of 3. the only page numbering that will equal the difference is page 1 plus 2. this gives a sum of 3. not too complicated despite the 'red herring' in the 'details and assumptions' that states that: the numbering begins with 1 on the front of the first page. the "remaining pages" include the pages BEFORE and after the page torn out, implying that the missing page was numbered 3 or greater since there were no pages before page 1.

Finding $\sum\limits_{n=1}^{32} x$ gives 528, which is 3 more than the remaining sum. The two pages numbers that were taken out were one and two, meaning $\boxed{3}$ is the correct answer.

Let n be the number of pages where n is any positive integer. The sum of the numbers on any page is 4n-1. The sum of all pages is the sum of all values from 1 to n for (4n-1). (I regret the text notation limits - pun intended). Let x be the number of the missing page. The sum of all pages in the magazine is 525 + 4x-1 or 524 + 4x. The only value that satisfies both equations is n = 16 (Sum from 1 to n = 528) and x = 1 (4x-1 = 3). Their are 16 pages (two sided, single sheet per page) in the magazine and the first page is missing.

I know this is very dumb, but I used a calculator and did: 1+2+3+4...+32+33=528. The different between 525 and 528 is 3!! (I know, it's the dumbest way to figure this out, and no it didn't take me long to do the calculator method.)

Suppose the document had nn pages. The sum of the page numbers is n(n+1)2.n(n+1)2. Suppose the removed page had numbers xx and xx+1+1with m odd. The removed page had a total of 2x+1.2x+1. Therefore,

n(n+1)2−2x+1=525,n(n+1)2−2x+1=525, or in frictionless form n2+n−4x−1048=0.n2+n−4x−1048=0. So 16x+420916x+4209 must be a perfect square, say m2m2. The only value of x which does the trick is 1. So the sum of the removed pages is 3.

Note then, that n=32.

We solve for coefficients of a quadratic using Method of Differences and obtain an expression for the sum of first n pages: $S(n)=(n+1)(2n+3)$ . The largest solution for $(n+1)(2n+3)=525$ is about $15$ . Plugging in $15$ gives $(15+1)(30+3)=528$ . Sum on all pages is odd. $528-3=525$ hence the sum of the page numbers on the page torn out is $3$ .

For me with my limited math skills It was a matter of attacking the problem from the right "angle"(or side to more precise)..:) My first attempt was to find the number of remaining pages (sum 525),assuming the last page was torn out.Soon i realised there was no way to arrive at 525 by this assumption.so i tried to assume the torn out page was on the beginning of the book,and voila!!!! I know ,not very mathematical but thats all i ve got :)

I went the dirty way. Found out the closest sum upto 525 which was sum of n=32 would be 528. Just add the sum of page 1 & 2 which gives you the answer. But the ideal way would be how Andy solved it with the equation.

Let $R_k$ be the sum of the recto (front side) page numbers on the first $k$ sheets of the book, and $V_k$ be the sum of the corresponding verso page numbers.

The verso page number of sheet $k$ is $2k$ , so $V_k = 2(1+...+k) = k(k+1) = k^2 + k$ .

Each recto page number of one less than its corresponding verso page, so $R_k = V_k - k = k^2$ .

So the page numbers of the first $k$ sheets sum to $N_k = R_k + V_k = 2k^2 + k$ .

We need to find $k$ such that $N_k > 525$ and $N_k - 525$ is odd (since a single sheet's page numbers always sum to an odd number). Since $N_k$ is even iff $k$ is even, we need the smallest even $k$ such that $N_k$ > 525.

Solving $x^2 + x = 525$ yields $x \approx 15.95$ (ignoring the negative root). So our our book has $k = 16$ sheets in total.

The missing sheet's page numbers sum to $N_k - 525 = 528 - 525 = \boxed{3}$ . (So it is sheet #1.)

Note: Going beyond 16 sheets won't work. If we suppose two more sheets (since $k$ must be even), we'd have $N_{18} - 525 = 141$ which would correspond to sheet #35.5, which is not only not a sheet number, but is nearly twice the size of the book. The gap between the required missing sheet number and the size of the book only grows with larger books, so even for book sizes that yield an integral missing sheet number, that sheet number will be larger than the book size. (E.g. a 20-page book would need page #69 ripped out.)