RMO 2009
Algebra
Level
4
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
Moderator note:
Great approach. Detailed solution that provides a step-by-step explanation.
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.