Question

In a book with page numbers from 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

Answer 1

its hard very

Answer 2

Math error. xD

Answer 3

There could really be a lot of answers to this. The sum off all the numbers from 1 to 100 is 5050, so we are missing a sum of 101. This could be just two numbers (50 and 51) or it could be 3 (48, 52, and 1), or any number of different answers.

Answer 4

actually, i think you are on to something without realizing it O.o if you tear out a page, you are tearing out 2 numbers at the same time. Which makes your case of 3 pages torn out impossible. Theres no way you could page 48 without tearing page 47, 52 without 51, so on and so forth. im not saying i have the answer, but this should lead to it.

Answer 5

pg 50 should be attached to page 49, so thats not it...

Answer 6

it may be that pga1 is on left side or right side

Answer 7

so i think there wud be two solutions for this

Answer 8

i want to think that most books start with pg 1 on the "right side" of the book. i dont know lol, this is just ideas.

Answer 9

yeah

Answer 10

but in mathseveything is possible LOL

Answer 11

Good call joemath - agreed.

Answer 12

i believe 3 pages are torn out. let the page numbers of these 3 pages be m, n, and l.
then the numbers than are torn out are m, m+1, n, n+1, l, l+1 which adds up to 2(m+n+1)+3. This should be 101.

Answer 13

Now, if we want integer solutions, there has to be an odd number of pages torn out. if there is an even number torn out, we would get something like:
2(m+n)+2 = 101
2(m+n) = 99
and that doesnt have integer solutions.

Answer 14

im trying to see if i can find an example where 5 pages makes a solution. for 3 pages i have: tear out 3 and 4, 21 and 22, 25 and 26

Answer 15

5 pages cant work, im gonna stick with my answer of three pages lol. i would like to stay and work on it more, but i have to get ready for school :( a very interesting problem indeed!

Answer 16

ok

Answer 17

THANKS for your time

Answer 18

yeah, ive confirmed it, it has to be 3, i'll be glad to post a solution later today if you'd like, but i gotta run >.<

Answer 19

and thank you jabberwock, i would have never seen it had you not said what you said lol

Answer 20

Thanks joemath, I'm glad you responded. I might give a similar problem to my students now.