1.

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?

Answer»

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< k< k<………<.kr . Then we see that

4(k1 + k+……….+ kr) – r ≥ 4 (k1 + k2 +……..….+ k7) – 7 ≥ 4 (1 + 2 +………..+ 7) - 7

=4 x 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.



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