InterviewSolution
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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? |
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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< k2 < k3 <………<.kr . Then we see that 4(k1 + k2 +……….+ 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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