A book has 100 pages numbered from 1 to 100. Some of its pages are

The sum of the numbers on the full book is:

N(N+1)/2 = 50*101 = 5050

Since the total of untorn pages is 4949, the sum of the torn pages is:

5050-4949 = 101

Now, each page has two numbers, so for every page torn, two numbers are removed from the total, an odd and an even number.

If 1 page is torn, the two page numbers can be represented by:

2K+1 and 2K+2, their sum being:

4k+3 = 101 or 4k=98. Since 98 isn't divisible by 4, more than one page must be torn.

If two pages are torn their numbers can be represented by:

2k+1,2k+2,2L+1 and 2L+2 and their sum:

4(k+L)+6 = 101 or

4(k+L) = 95. Since 95 isn't divisible by 4, more than two pages must be torn.

How about 3 pages ? As above:

4(k+L+m) + 9 = 101 or:

4(k+L+m) = 92 or

K+L+M = 23

So, any combination of three different numbers adding to 23 will work.

For example:

K=12, l=8, m=3, which yield the following page numbers:

K:25 and 26 = 51

L:17 and 18 = 35

M: 7 and 8 = 15

Totaling 101


So three pages is the correct answer



Posted from my mobile device

Thank you so much. My bad, how come I missed this silly thing, that one page will include both the number. Thanks

If the pages are stitched together in the book, then the last page and the first page should be stitched together. This pattern will follow for all the rest of the pages. So, if we tear the first page, the last page also gets torn automatically. If we consider this, then first page is numbered as 1 and the last page as 100. If we tear the page, it gives us 101. 2 can be a probable answer as well if we think like this.
What am I missing here ?

What happen if I tore only 1 page that is 50 and 51
Would t not be correct as well?

The first sheet has the number 1 on the front side and number 2 on the opposite side as you flip the page.

So page 50 is on the same sheet as page 49, and page 51 is on the same sheet as page 52.

So, two sheets comprise pages 50 and 51, not one.

So, if you tear the sheet containing page 50 you are also eliminating 49 from the count and similarly 52 is eliminated from the count when tearing the sheet containing 51.

These numbers add:

49+50+51+52 = 202

Twice the target of 101.

Posted from my mobile device

Every sheet has two numbers.

The first has page numbers 1 and 2.

The sheet can't be torn and only eliminate page 1.

So your example would actually eliminate page numbers 1 and 2 as well as page numbers 99 and 100, for a total of:

1+2+99+100 = 202.

Twice the 101 objective.

Posted from my mobile device

If I tear a page which contains page number 50 on front and 51 on back. It means only 1 page is remove and the sum is also 50+51 that is 101.

It is also possible right.

No.

The first page has a 1 on the front and a 2 on the back.

Pages can't have an even number on the front.

I suggest opening a book and see for yourself.

Posted from my mobile device

Hey Bunuel

The question starts with "A book has 100 pages" and ends with asking "how many pages are torn". So, doesn't this mean that one page has only 1 numbering given pages are numbered 1 to 100. If if it is numbered on both sides then it should have 50 pages (2 numbers on each side of the page). This seems confusing.