There are a certain number of pages in a book. Bob tore a certain page out of the book and later found that the average of the remaining page numbers is $46 \frac{10}{13}$. What were the page numbers that Bob had torn ?
Please let me know where am I going wrong with my approach :-
Since the average of the remaining "page numbers" has 13 in the denominator , I can say that the remaining number of "page numbers" have to be a multiple of 13
So the remaining number of "page numbers" can be 13, 26, 39, 52, 65, 78, 91, 104.....
Accordingly the original number of "page numbers" in the book could be 15, 28, 41, 54, 67, 80, 93, 106...
Now, Sum of remaining "page numbers" + sum of the torn "page numbers" = sum of total "page numbers" initially
I started making cases now,
Case 1:- When original number of "page numbers" were 15
So it would have "page numbers" from 1 till 15, therefore total of all "page numbers" would be (15*16)/2 = 120
Sum of remaining "page numbers" = $46 \frac{10}{13}$ * 13
Let sum of torn pages = x
x + ($46 \frac{10}{13}$ * 13) = 120
which gives x to be a negative number
Case 2:- When original number of "page numbers" were 28
So it would have "page numbers" from 1 till 28, therefore total of all "page numbers" would be (28*29)/2 = 406
Sum of remaining "page numbers" = $46 \frac{10}{13}$ * 26
Let sum of torn pages = x
x + ($46 \frac{10}{13}$ * 26) = 406
x is still negative
Case 3:- When original number of "page numbers" were 41
So it would have "page numbers" from 1 till 28, therefore total of all "page numbers" would be (41*42)/2 = 861
Sum of remaining "page numbers" = $46 \frac{10}{13}$ * 39
Let sum of torn pages = x
x + ($46 \frac{10}{13}$ * 39) = 861
x is still negative :(
Case 4:- When original number of "page numbers" were 54
So it would have "page numbers" from 1 till 54, therefore total of all "page numbers" would be (54*55)/2 = 1485
Sum of remaining "page numbers" = $46 \frac{10}{13}$ * 52
Let sum of torn pages = x
x + ($46 \frac{10}{13}$ * 52) = 1485
x<0
Case 5:- When original number of "page numbers" were 67
So it would have "page numbers" from 1 till 67, therefore total of all "page numbers" would be (67*68)/2 = 2278
Sum of remaining "page numbers" = $46 \frac{10}{13}$ * 65
Let sum of torn pages = x
x + ($46 \frac{10}{13}$ * 65) = 2278
x<0
Case 6:- When original number of "page numbers" were 80
So it would have "page numbers" from 1 till 54, therefore total of all "page numbers" would be (80*81)/2 = 3240
Sum of remaining "page numbers" = $46 \frac{10}{13}$ * 78
Let sum of torn pages = x
x + ($46 \frac{10}{13}$ * 78) = 3240
x<0
Case 7:- When original number of "page numbers" were 93
So it would have "page numbers" from 1 till 93, therefore total of all "page numbers" would be (93*94)/2 = 4371
Sum of remaining "page numbers" = $46 \frac{10}{13}$ * 91
Let sum of torn pages = x
x + ($46 \frac{10}{13}$ * 91) = 4371
x=115 (:O)
now is it that 2 consecutive page numbers m+m+1=115
one page number will be 57 and other 58 and both of them even lie in the range of 93 "page numbers" as well
Per the comments on the original question, the issue with the logic appears to be some confusion about pages vs page numbers.
I think there's a secondary issue of going for cases a bit too early - it's worth thinking heuristically about the problem first to get a feel for what sensible answers might look like.
Before going into cases, it's worth trying to get a feel for the problem. Ripping out a single page won't change the average by much. What's the total of all page numbers in a book with page numbers up to $n$? It's the $n^{\text{th}}$ triangular number, ie $\frac12 n(n+1)$, so the average is $\frac12 (n+1)$. In other words, without really doing any work, we know the page numbers should go up to around $100$.
You can just do cases here and find the solution, but in case it's of interest I've put the rest of a solution below (spoiler alert!).
Say the torn out page has numbers $k$ and $k+1$ on it. This, with the given information, gives us $$\frac{\frac12 n(n+1)-2k-1}{n-2} = 46\frac{10}{13}=\frac{608}{13}$$
As you noticed, this means $n-2$ is a multiple of $13$; let's say $n=13m+2$. Now, substituting and tidying up a bit, the above expression becomes $$ 169 m^2 - 1151 m + 4 = 4k$$
We need $0 < k < n = 13m+2$, so that $$ 0 < 169 m^2 - 1151 m + 4 < 52m+8$$
Solving this, we find that $6.8 < m < 7.1$, so the only possibility is $m=7$; and this does indeed work. We find the original book's page numbers went up to $93$, and the page with numbers $57$ and $58$ was removed.
If the denominator in the problem had been something less convenient, this approach would still work on the equation for $k$ in terms of $n$, but there may have been a larger range of cases to check.