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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

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2 Answers 2

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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, $n=3.$

Edit

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).)

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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