Problem of the torn pages in a book

[agnibho]

Homework Statement

A page is torn from a book with pages numbered 1 to 100. The sum of the page numbers(both sides) of the remaining pages is 4949. What is the sum of the page numbers on the torn page? If possible, can you find the page numbers of that specific page?2.An attempt at the solution
Suppose r pages of the book are torn off. Note that the page numbers
on both the sides of a page are of the form 2k − 1 and 2k, and their sum is 4k − 1.
The sum of the numbers on the torn pages must be of the form
4k1 − 1 + 4k2 − 1 + · · · + 4kr − 1 = 4(k1 + k2 + · · · + kr) − r.
The sum of the numbers of all the pages in the untorn book is
1 + 2 + 3 + · · · + 100 = 5050.
Hence the sum of the numbers on the torn pages is
5050 − 4949 = 101.
We therefore have
4(k1 + k2 + · · · + kr) − r = 101.
This shows that r ≡ 3 (mod 4). Thus r = 4l + 3 for some l ≥ 0.
Suppose r ≥ 7, and suppose k1 < k2 < k3 < · · · < kr. Then we see that
4(k1 + k2 + · · · + kr) − r ≥ 4(k1 + k2 + · · · + k7) − 7
≥ 4(1 + 2 + · · · + 7) − 7
= 4 × 28 − 7 = 105 > 101.
Hence r = 3.
This leads to k1 + k2 + k3 = 26
Am I right?? Actually the answer's not given in my book. So I am a bit confused.

 

[LCKurtz]

Your problem states that one page is torn from the book, so it is much simpler than what you are working on. I think you will find it is page 50,51.

 

[phinds]

Homework Statement

A page is torn from a book with pages numbered 1 to 100. The sum of the page numbers(both sides) of the remaining pages is 4949. What is the sum of the page numbers on the torn page? If possible, can you find the page numbers of that specific page?


2.An attempt at the solution
Suppose r pages of the book are torn off. Note that the page numbers
on both the sides of a page are of the form 2k − 1 and 2k, and their sum is 4k − 1.
The sum of the numbers on the torn pages must be of the form
4k1 − 1 + 4k2 − 1 + · · · + 4kr − 1 = 4(k1 + k2 + · · · + kr) − r.
The sum of the numbers of all the pages in the untorn book is
1 + 2 + 3 + · · · + 100 = 5050.
Hence the sum of the numbers on the torn pages is
5050 − 4949 = 101.
We therefore have
4(k1 + k2 + · · · + kr) − r = 101.
This shows that r ≡ 3 (mod 4). Thus r = 4l + 3 for some l ≥ 0.
Suppose r ≥ 7, and suppose k1 < k2 < k3 < · · · < kr. Then we see that
4(k1 + k2 + · · · + kr) − r ≥ 4(k1 + k2 + · · · + k7) − 7
≥ 4(1 + 2 + · · · + 7) − 7
= 4 × 28 − 7 = 105 > 101.
Hence r = 3.
This leads to k1 + k2 + k3 = 26
Am I right?? Actually the answer's not given in my book. So I am a bit confused.

WHOA are you ever making this MUCH harder than it needs to be.

LCKurtz has it right, although it is unfortunate that he just spoonfed you the answer rather than offering you help in figuring out how to GET the answer so you would learn something rather than just having the answer.

 

[LCKurtz]

LCKurtz has it right, although it is unfortunate that he just spoonfed you the answer rather than offering you help in figuring out how to GET the answer so you would learn something rather than just having the answer.

Easy there. The point is I didn't give him help on the argument for two reasons.
1. I want him to work it out himself.
2. It is pretty apparent from his post that he will be able to solve it with no additional help.

 

[phinds]

Easy there. The point is I didn't give him help on the argument for two reasons.
1. I want him to work it out himself.
2. It is pretty apparent from his post that he will be able to solve it with no additional help.

I always think that "help on the argument" helps people learn and just giving the answer offers them to possiblity of learning nothing.

 

[DrewD]

most books don't have a page 50,51. There is a page 49,50 and 51,52. The problem does seem to be asking for a single page though, so this must be a self-published book. I think LCKurtz's assumption that agnibho could figure the easier answer out was justified. It looks like the solution that agnibho came up with is correct with the restriction that $k_n\geq0$

 

[SammyS]

I always think that "help on the argument" helps people learn and just giving the answer offers them to possiblity of learning nothing.

I have followed many, many of LCKurtz's posts. As a rule, he does not give out solutions. On the contrary, he gets quite upset with those who do simply give out solutions.

I'm sure he had good reason for the response he gave.

SammyS

 

[phinds]

I have followed many, many of LCKurtz's posts. As a rule, he does not give out solutions. On the contrary, he gets quite upset with those who do simply give out solutions.

I'm sure he had good reason for the response he gave.

SammyS

Glad to hear it. I too get upset when someone just blindly (it seems to me) gives the answer instead of giving actual help, which is why I got irate in this case.

 

[phinds]

most books don't have a page 50,51

DOH !



I didn't open my brain long enough even notice that aspect of the problem. Color me embarrased.

 

[agnibho]

Thanks for the suggestions. And, I will work out the sum again on my own and maybe post it here too. So, don't worry I won't blindly follow the answer, because if I do so, I won't be able to work out other similar problems which maybe of higher level. It's just that I wanted to know whether I was on the right track or not. Thanks again.