q

[lost oxide]

2

100 * 101 / 2 = 5050

that means we turn off two pages

[chrome schooner]

I agree

[lost oxide]

damn bro

the pattern for sum numbers from 1 to n is n * (n + 1) / 2

you can see that one from n , n + 1 is divisible by 2

and the answer is integer

[chrome schooner]

Dang I wasted my time when I could just have used that formula

[low lily]

Well look ||you tore a page, which means you should count the sum front and back||, and ||since an odd number appears, an odd number of pages were torn|| @dawn notch . Its late for me rn so these Hints should help you

Use only one hint at a time please. First one first

[pure eagle]

Of course, ||50-th and 51-th were torn off. Because 5050-50-51=4949.||

[low lily]

!nosols

[eager glen]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[low lily]

Please

[pure eagle]

but that's of course is not a complete solution

[novel cargo]

That's interesting cuz a book should start with page 1, that mean there can't be a page numbered 50 and 51

[pure eagle]

yep, you are right. So, there is a contradiction

[low lily]

I dont want to use !noans now lol

But it's wrong anyways

[pure eagle]

it depend's on interpretation

[low lily]

Books general start from 1

So 51 and 52 would be one page

And 49 and 50 another

I haven't see a book start from 0, have you?

[pure eagle]

in general. Are there books where number 1 is a backside of the cover?

[low lily]

I haven't seen thst

Not a general thing atleast

[pure eagle]

but it is easy to make the one of that kind

[tall junco]

they don't say pages numbered front and back

[low lily]

Let OP clarify then

[tall junco]

(tho this would make least sense)

[low lily]

Till there are no signs of life, we should leave it

[low lily]

Yeah

[low lily]

I dont think so people make such peculiar problems. The cover is separate is ~100% of books

[pure eagle]

I think that page numbering pattern is pretty common for books in Arabic and Hebrew.

[tall junco]

:mammaia

[pure eagle]

So, the answer may be 1 🙂

[novel cargo]

,calc 5050-11-12-15-16-23-24

[young marlin]

Result:

4949

[novel cargo]

or 3

[pure eagle]

yes, there are many ways to get 101 as a sum of distinct numbers. That gives many possible answers.

though if we turn on that limitation that we tear off both (2n-1)-th and 2n-th pages then it limits the answer a little

[dawn notch]

it has many solutions

[tall junco]

how many

[dawn notch]

the sum should be 101

[pure eagle]

so we need the number of ways to represent 101 as a sum of distinct numbers of type 4k-1

[novel cargo]

actually for first page start with 1 there're 2 solutions

But in both we torn 3 pages

so like, it's not exactly 2

[dawn notch]

You can tear any amount of pages other than 1

note that

[novel cargo]

Well I haven't done such case so

No idea if there're more than 2

[pure eagle]

yep, the only possible way to do that is to tear off exactly 3 pages.

Here are all 44 representations:
3 + 7 + 91
3 + 11 + 87
3 + 15 + 83
3 + 19 + 79
3 + 23 + 75
3 + 27 + 71
3 + 31 + 67
3 + 35 + 63
3 + 39 + 59
3 + 43 + 55
3 + 47 + 51
7 + 11 + 83
7 + 15 + 79
7 + 19 + 75
7 + 23 + 71
7 + 27 + 67
7 + 31 + 63
7 + 35 + 59
7 + 39 + 55
7 + 43 + 51
11 + 15 + 75
11 + 19 + 71
11 + 23 + 67
11 + 27 + 63
11 + 31 + 59
11 + 35 + 55
11 + 39 + 51
11 + 43 + 47
15 + 19 + 67
15 + 23 + 63
15 + 27 + 59
15 + 31 + 55
15 + 35 + 51
15 + 39 + 47
19 + 23 + 59
19 + 27 + 55
19 + 31 + 51
19 + 35 + 47
19 + 39 + 43
23 + 27 + 51
23 + 31 + 47
23 + 35 + 43
27 + 31 + 43
27 + 35 + 39

[tall junco]

what do the 3 numbers represent?

sum of page numbers on a page?

[pure eagle]

yep

say 23+31+47 means 11,12 + 15,16 + 23,24

since 101 is of type 4k+1 then we need either 3, or 7 or 11 pages. But anything greater than 3 exceeds 101 (probably)

[dawn notch]

Pls note that it can be any amount of pages that can sum 101

[novel cargo]

only 3 satisfy

[tall junco]

wdym by any amount

[pure eagle]

the fact that other than 3 means 7 or more, makes it impossible

[dawn notch]

the number of pages that you can tear

[tall junco]

by parity 2 is not possible

so it leaves with 3

[dawn notch]

How?

[tall junco]

bieng said you need to clarify, this

[novel cargo]

Any number that's not 3 is not possible tbh in our case

[dawn notch]

Singular page

[tall junco]

when will you mention

[dawn notch]

Only front

[pure eagle]

Since 1+2+...+100=5050 and 5050-4949=101, the total sum of numbers on torn off pages is 101.

101 is a number of the form 4k+1, but each torn off page gives 4n-1 as a sum of numbers (2n-1, 2n) on it. So, the number of terms in 101 (aka the number of torn off pages) should be 3, 7, 11 ...
but 3 + 7 + 11 + 15 + 19 + 23 + 27=105 which is >101
so, only 3 pages could have been torn off
and we see that it is possible. Say, (1,2)+(3,4)+(45,46)= 101

[tall junco]

if we consider this, we need solutions for sum xi = 101

where number of terms increase

[dawn notch]

That's what the question says

[novel cargo]

aww, that makes it less interesting

[dawn notch]

That also makes my calculations harder

[pure eagle]

But that's only true for western books. For books in Arabic and Hebrew the sum of the number on one pege and on it's backside is of the type 4n+1. So, only 5 pages could be torn off, (because 1 is impossible)
say (2,3)+(4,5)+(6,7)+(8,9)+(23,24)=101

[novel cargo]

Then there're many cases, I mean it can be 2-3... pages

Asking "How many pages were torn off?" doesn't make sense here

[novel cargo]

If instead the question is, in how many way we can do such thing then basically we have to find numbers of solutions for this

[tall junco]

i feel this is made up

[novel cargo]

that's not easy tho, we might have to consider recursion

[wide remnant]

damn so many comments

this question can be done by bounding I think

[tall junco]

hmm..

[pure eagle]

here is a complete solution #1452325283116879952 message

[wide remnant]

hmm

[dawn notch]

Bro it's the question on RMO(Regional Math Olympiad)

[tall junco]

ic

which year

[dawn notch]

2009

[wide remnant]

damn that's useless

[dawn notch]

Absolute trash

[novel cargo]

It equal to number of partitions of a set with 101 elements such that no 2 sets are equal in size

[wide remnant]

how the hell is this related to university level math tho 💀

[tall junco]

broski they say they are numbered front and back

[novel cargo]

That's basically what we were doing, right?

[dawn notch]

I feel like a loser

[tall junco]

yeah,what previously was mentioned

[tall junco]

this

[dawn notch]

Should've seen the solution instead of asking.

Now it has 100+comments

[wide remnant]

that's a pretty cool solution

mod 4 and it's insta solved

[tall junco]

hmm.. but now if we look at other situation

[wide remnant]

altho in my opinion I think the problem was more about reading comprehension than math in itself

[tall junco]

that is how they set these up