I was solving some old olympiad problems and I got that one. I m stuck at it.

"In a book with page numbers 1 to 100,some pages are torn off. The sum of the numbers on the remaining pages is 4949. How many pages are torn off??"

I tried to the summation formula(sum of all the pages is sum of all the natural numbers from 1 to 100,since the pages goes from 1 to 100) and then try to eliminate some numbers (by hit and trial) to see whether the sum comes out to be 4949 or not but I wasn't so lucky. Hoping for help. Any suggestion is heartily welcome


2 Answers 2


We have that $1+\cdots+100=5050.$ So, if the remaining pages sum $4949$ the pages which are not in the book add up $101.$ Moreover, if $2a-1$ is not in the book then $2a$ is also not in the book. They add up $4a-1.$ It's not possible to torn out only a page because $4a-1=101$ has no integer solution. If we turn out two pages we have to solve $4a-1+4b-1=101$ which also doesn't have integer solution. So, assume $n$ pages are torn out. We have $$4a_1-1+\cdots+4a_n-1=101 (\iff 4(a_1+\cdots+a_n)=101+n).$$ Thus the number of pages can be $3,7,11, \dots$ because $101+n$ must be a multiple of $4.$ Now, note that if $n\ge 7$ then $$\dfrac{n(n+1)}{2}=1+\cdots+n\le a_1+\cdots+a_n=\dfrac{101+n}{4}< \dfrac{n(n+1)}{2}$$ gives a contradiction. So, $n=3.$


Let's prove of the last inequality for $n\ge 7:$ $$\dfrac{101+n}{4}< \dfrac{n(n+1)}{2}\iff 2n^2+2n>101+n\iff 2n^2+n>101.$$ Now, $$n\ge 7\implies 2n^2+n\ge 98+7=105>101,$$ and we are done.

Edit 2

In order to determine the pages we have to solve $$a_1+a_2+a_3=26.$$ Then, the pages are $2a_1-1,2a_1;$ $2a_2-1,2a_2;$ and $2a_3-1,2a_3.$

Now, solving $a_1+a_2+a_3=26$ is to get the partitions of $26$ into distinct parts (that is, $a_1,a_2$ and $a_3$ are different). According to Wolfram Alpha there are $165$ solutions. (See https://www.wolframalpha.com/input/?i=PartitionsQ(26).)

  • $\begingroup$ what you did in the last step??Can u please elaborate it?? $\endgroup$ Nov 8, 2016 at 14:12
  • $\begingroup$ I too think the last step is a little confusing. Each of the comparisons has a different reason, some more obvious than others. The $<$ at the end is the only part that depends on the value of $n$. It might be easier if this were spelled out in a little more detail. $\endgroup$
    – David K
    Nov 8, 2016 at 14:14
  • $\begingroup$ Is it clearer now? $\endgroup$
    – mfl
    Nov 8, 2016 at 14:27
  • $\begingroup$ Yeah,pretty much. $\endgroup$ Nov 8, 2016 at 14:29
  • $\begingroup$ It depends on you. But the answer is not unique. Pages 1-2,3-4,45-46 and 1-2, 5-6, 43-44 are solutions. There are more solutions. $\endgroup$
    – mfl
    Nov 8, 2016 at 14:35

Start by summing up all the pages from $1$ to $100$ to see how much it would add up if no pages are torn. Then it should become clear.

  • $\begingroup$ You also have to be aware of the fact that tearing out page $i$ also removes page $i+1$ or page $i-1$, depending on whether $i$ is even or odd. $\endgroup$ Nov 8, 2016 at 13:55
  • $\begingroup$ @RickDecker that is right but in no way goes against anything I said. I only pointed him in the right direction. $\endgroup$
    – RGS
    Nov 8, 2016 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.