TITLE: Help me solve my father's riddle and get my book back
QUESTION [78 upvotes]: My father is a mathteacher and as such he regards asking tricky questions and playing mathematical pranks on me once in a while as part of his parental duty.
So today before leaving home he sneaked into my room and took the book I am currently reading!
The book is quite old and damaged with one or two pages torn out and as I checked my phone in the morning I find a message along the lines of this:

[A picture of him proudly grinning and holding a torn out page in his hand]
Dear Levix, if you want to know where your book lies then tell me: What page am I holding when the sum of all remaining page numbers (without those 2 he is holding) is equal to $81707$? :)

Can anybody provide any advice? (it would be awesome if we could find a general solution to stick it to the man for good. ;) )
Update: First, I want to thank you all for your kind effort and for helping me out so rapidly! I enjoyed your intelligible answers so much that I couldn't resist to use this knowledge against him :) The final response I gave was that If the sum of all remaining page numbers had been my birthday than the last 2 digits + 10 (32 41, 32 42) would have added up to the 
page numbers of the turn out page he was holding. I not only got my book back - I also received a great big hug. So thank you! 
(Pluspoints if you can calculate my birthday)

REPLY [7 votes]: Here's a simple solution to the problem using a bit of programming.

Let remainingPageSum equal 81707.
Let a page number counter pageNumber equal 1.
Let totalPageSum equal pageNumber.
While totalPageSum is less than remainingPageSum:

Increment pageNumber by 1, and
Increment totalPageSum by pageNumber.


Now you have totalPageSum equal to 81810. Subtract remainingPageSum from totalPageSum to get the sum of the removed page numbers, 103. The two consecutive page numbers having a sum of 103 will be floor(103 / 2) and ceil(103 / 2), or 51 and 52.
Test it out here.