jmc

Question

The pages of a book are numbered 1_^ through n_^. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of 1986_^. What was the number of the page that was added twice?

Accepted Answer

If all the page numbers were added just once, the sum would be 1 + 2 + + n = n(n+1)2.But one of the page numbers was added twice, so the interval of possible values for the incorrect sum is [n(n+1)2 + 1, n(n+1)2 + n]. We are given that the incorrect sum is 1986, so we must have n(n+1)2 + 1 1986 n(n+1)2 + n.We look for the value of n which satisfies this inequality chain. We have n(n+1)2 1986 2000, so n(n+1) n^2 4000, and n 4000 63. For n = 63, we have n(n+1)2 = 2016, which is too large. For n=62, we have n(n+1)2 = 1953, which works, because 1953 + 1 1986 1953 + 62.Then, the page number that was added twice must be 1986 - 1953 = 33.