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 +………….+4k_r – 1 = 4(k₁ + k₂ +…….+ k_r) – 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 +…….+ k_r) – 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 <………<.k_r . Then we see that

4(k₁ + k₂ +……….+ k_r) – 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.