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

100 points -

Answer 2

a complete book shud have
\[s_{100}=\frac{100(100+1)}{2}=5050\]
right

Answer 3

sum of 1st hundered nos. = 5050
5050-4949 = 101

Now note that for any page, odd no. comes first on the right side and even no. comes at the left.
like page nos will be (1,2) -> 1 on right, and when you turn over, 2 will be on left.
then there will be (3,4) -> 3 on right, 4 on left.
.
.
likewise.
for 1 page, a+ a+1 = 101
or a=50
but a should be odd, hence wrong ans.
for 2 pages,
2a+1 + 2b+1 = 101
a+b = 99/2 -->not possible
for 3 pages,
a+b+c = 49

a,b,c all are odd
this is possible
hence 3 can be an answer.

infact, 3 should be the answer.

Answer 4

if i tore page 1 and 100=101
5050-101=4049

Answer 5

you can not tear 1 and 100

Answer 6

when you tear one page you remove 2 numbers

Answer 7

yes thanx for that clearance and 3 is correct answer ,why does @shubhamsrg suppose that 3 has to be correct and no oher answer thereof,

Answer 8

I dont have a concrete proof for that, i just said said by hit and trial, but I am trying to think of a reason

Answer 9

I assume the pages being torn off must be sequencial?

Answer 10

how many pages were there on book originally? i could be 50 or 52 ... assuming there are two blank pages.

Answer 11

blank or not its still a page so we count it

Answer 12

if that is the case, then the sum of the pages torn off wil be given by:\[\frac{n}{2}(2a+(n-1)d)=101\]where n is the number of pages torn off, 'a' is the first page number, and d=1.
you coud then make use of the fact that 101 is prime

Answer 13

I don't think we only tear off consecutive pages.

Answer 14

yeah we cud tear numbers that are not consecutive,but it seems like 3 is a unique solution anyway

Answer 15

on the solution it just says 3 and comment "unique"

Answer 16

why couldn't it be one page torn off - with 50 on one side and 51 on the other?

Answer 17

i think it depends on where 1 stars, 50 and 51 may not be on same page.

Answer 18

probably cos they are not on the same poage

Answer 19

how do we verify that

Answer 20

there doesn't seem to be enough information given to assert that

Answer 21

for n pages torn,
2(a_1+a_2+a_3 ...a_n) + n = 101
(a_1+a_2+a_3..a_n) = (101-n)/2

a_1,a_2,a_3..a_n all are odd. n also has to be odd so as to make RHS an integer.

Another condition is RHS should be odd integer.

n=3,7,11.. satisfy this criteria.
for n=7,
we have a_1 + a_2 .. a_7 = 47

sum of first 7 odd numbers is 49 i.e. greater that 47
hence anything greater than 3 should not be solution .

Answer 22

every page should have 2 numbers on either side, hence I;d say we can conclude (49,50) will be one leaf and (50,51) the other

Answer 23

sorry, (49,50) one and (51,52) other

Answer 24

but 49+50+51+52 does not equal 101?

Answer 25

if 1 starts on front page, then 50 and 51 are not on same page. if 1 starts on back of first page then 50 and 51 are on same page.
since the sum is odd, it's good to look for odd number of pages.

Answer 26

right, it isn't, so (49,50) and (51,52) is not a solution.

Answer 27

@experimentX I agree - so why isn't 1 a solution? i.e. the page with 50 on one side and 51 on the other

Answer 28

bcos 1 starts on front page, then 50 and 51 are not on same page

Answer 29

yeah otherwise front page wont have an index

Answer 30

so what pages are this ????

Answer 31

i mean what pages are torn?

Answer 32

ok ... since you clarified i suppose it's better to look the solution of the form
http://www.wolframalpha.com/input/?i=integer+solutions+%282n%2B1%29%2B%282n%2B2%29%2B%282m%2B1%29%2B%282m%2B2%29%2B%282o%2B1%29%2B%282o%2B2%29+%3D+101

Answer 33

so there are many possibilities for a combination of 3 pages?

Answer 34

right, many.

Answer 35

should be many I mean

Answer 36

I see - so we can assert that 3 pages have been torn off, but which 3 cannot be determined from the information given.

Answer 37

you mean which are obviusly not torn?

Answer 38

is it can or cannot be dertemined

Answer 39

also consider if 5 pages might have been torn off, i haven't checked it out.

Answer 40

in general, i would be advisible to look for odd number of pages.

Answer 41

but I thought my hypothesis was a good proof that more than 3 pages is not possible :|

Answer 42

I agree with @shubhamsrg

Answer 43

thanks a lot guys for all the input ,you guys are really great problem solvers
...i am out of medals...

Answer 44

@Jonask the question asks "how many pages are torn off" - so answer is 3. We cannot determine exactly which 3 but we do know that exactly 3 pages were torn off. See the reasoning outlined by @shubhamsrg above.

Answer 45

yes the question is completlty answered,i understand the solution completly

Answer 46

can we determine how many solutions are possible

Answer 47

I think it would be the number of solutions for:\[a+b+c=49\]where a, b and c are all odd integers

Answer 48

yeah 3 variables so many !

Answer 49

:)

Answer 50

i see number theory here i think we shud halt here

Answer 51

you can certainly solve it by brute force as the range is small :)

Answer 52

its a permutation and combination thing then

Answer 53

its surely less than 253, not sure how much less

Answer 54

should be of the form a+b+c = 23 where a>0, b>0, c>0 and \( a \neq b \neq c \)

Answer 55

if we remove condition for a not equal b not equal c,
then 25C2 or300 solutions we have
but it surely has to be less than that coz of a not equals b not equals c

I said 253 after assuming a>1,b>1,c>1
but that was wrong assumption, hmm..

Answer 56

let a=1, b+c=22 => 19 solutions
a=2, b+c= 21=> 19 solutions
a=3, b+c= 20=> 18 solutions
..
..

Answer 57

aha
nice.

Answer 58

yes - nice reduction @experimentX :)

Answer 59

thanks :)
still i haven't got or thought why more than three pages can't be torn off.

Answer 60

I'll simply copy-paste from above
for n pages torn,
2(a_1+a_2+a_3 ...a_n) + n = 101
(a_1+a_2+a_3..a_n) = (101-n)/2

a_1,a_2,a_3..a_n all are odd. n also has to be odd so as to make RHS an integer.

Another condition is RHS should be odd integer.

n=3,7,11.. satisfy this criteria.
for n=7,
we have a_1 + a_2 .. a_7 = 47

sum of first 7 odd numbers is 49 i.e. greater that 47
hence anything greater than 3 should not be solution.

Answer 61

@shubhamsrg explained it nicely above...

Answer 62

and now repeated just above :)

Answer 63

i see :)

Answer 64

i see :) well explained ,,,