And here is an easy one
Number Theory
Level
pending
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?
The answer is 3.
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1 solution
Suppose r pages of the book are torn off. Note that the page numbers on both the sides of a page are of the form 2k −1 and 2k, and their sum is 4k −1. The sum of the numbers on the torn pages must be of the form 4k1 − 1 + 4k2 − 1 + ··· + 4kr − 1 = 4(k1 + k2 + ··· + kr) − r. The sum of the numbers of all the pages in the untorn book is 1 + 2 + 3 + ··· + 100 = 5050. Hence the sum of the numbers on the torn pages is 5050 − 4949 = 101. We therefore have 4(k1 + k2 + ··· + kr) − r = 101. This shows that r ≡ 3 (mod 4). Thus r = 4l + 3 for some l ≥ 0. Suppose r ≥ 7, and suppose k1 < k2 < k3 < ··· < kr. Then we see that 4(k1 + k2 + ··· + kr) − r ≥ 4(k1 + k2 + ··· + k7) − 7 ≥ 4(1 + 2 + ··· + 7) − 7 = 4 × 28 − 7 = 105 > 101. Hence r = 3. This leads to k1 + k2 + k3 = 26 and one can choose distinct positive integers k1,k2,k3 in several ways.