Arithmetic progression problem (I guess)

Darya

Junior Member
Joined
Jan 17, 2020
Messages
154
So I've got this problem: A sheet was torn from a book (which is e.g. pages 22,23). The sum of all the remaining pages is 7495. How many pages were there originally in the book if they start from 1?

I've tried to fit that in an arithmetic progression formula but then I got too many variables. Any ideas on how to start better?
 
What's the formula you are trying to use and how did you use it?
 
I would let \(P\) be the number of pages originally in the book and write:

[MATH]0<\frac{P(P+1)}{2}-7495<2P[/MATH]
You should find two integer solutions, only one of which has the difference in the middle being odd (why do we need this difference to be odd?).
 
Tried to use a formula of sum
((1+n)/2)n=7495 although it doesn't seem correct at all
This formula applies to a valid sequence. Without 22 and 23 the sequence is not a valid arithmetic sequence. How do we make it valid?
Edit: disregard, I didn't notice that 22 and 23 were an example.
 
So I've got this problem: A sheet was torn from a book (which is e.g. pages 22,23). The sum of all the remaining pages is 7495. How many pages were there originally in the book if they start from 1?
This is a faulty question.
Pickup any book or magazine a single sheet is numbered with two integers, one odd & one even.
The odd is less than the even. Whoever wrote the has not dealt much with books.
 
Your thought process matches mine - AP.

So how many pages did the original (intact) book had?
 
This is a faulty question.
Pickup any book or magazine a single sheet is numbered with two integers, one odd & one even.
The odd is less than the even. Whoever wrote the has not dealt much with books.
What if the last page was torn - and it was printed (and counted) only on one side (i.e. the last printed page was an odd page).
 
This is a faulty question.
Pickup any book or magazine a single sheet is numbered with two integers, one odd & one even.
The odd is less than the even. Whoever wrote the has not dealt much with books.
Okay. Sure, let's do that.


A few pages ago, there was a single sheet containing the pages 22 and 23.
 
What if the last page was torn - and it was printed (and counted) only on one side (i.e. the last printed page was an odd page).
And that is often the case. But in this case there is no indication of that.
 
Tried to use a formula of sum
((1+n)/2)n=7495 although it doesn't seem correct at all
Of course this formula will not work (unless the last sheet is missing) as you missed two numbers in the sum. Is there a sum using that formula that is just above 7495 and the difference is odd. You can try to figure it out that way.
 
Okay. Sure, let's do that.


A few pages ago, there was a single sheet containing the pages 22 and 23.
Okay, I'm curious. Where did you find this defective book (which I presume is not in your possession, or you would have shown those pages)?

According to https://en.wikipedia.org/wiki/Page_numbering,

Even numbers usually appear on verso (left-hand) pages, while odd numbers appear on recto (right-hand) pages. In the printing industry, in cases where odd numbers appear on verso pages and even numbers on recto pages, this is referred to as non-traditional folios

So there are some books that don't follow the convention, but they should be rare (and in some cases it may be a mistake).

In any case, the example isn't necessarily meant to be real. The answer does fit the usual expectation.
 
I would let \(P\) be the number of pages originally in the book and write:

[MATH]0<\frac{P(P+1)}{2}-7495<2P[/MATH]
You should find two integer solutions, only one of which has the difference in the middle being odd (why do we need this difference to be odd?).
Thank you soo much! I got the right answer... But could you please also explain the logic behind this equation in short? I can't quite get it...
 
Thank you soo much! I got the right answer... But could you please also explain the logic behind this equation in short? I can't quite get it...

If P is the number of pages and the missing pages are A and B:

[MATH]\frac{P(P+1)}{2} = 7495 + A + B[/MATH]
Since A + B < 2P:

[MATH]\frac{P(P+1)}{2} < 7495 + 2P[/MATH]
 
If P is the number of pages and the missing pages are A and B:

[MATH]\frac{P(P+1)}{2} = 7495 + A + B[/MATH]
Since A + B < 2P:

[MATH]\frac{P(P+1)}{2} < 7495 + 2P[/MATH]
God bless you!!! I'm really grateful
 
Top