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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Solution

Suppose $r$ pages of the book are tron off.
Note that the page numbers on both the sides of a page are of the form $2 k - 1$ and $2 k$ , and their sum is $4 k - 1$ .
The sum of the numbers on the torn pages must be of the form $4 k_{1} - 1 + 4 k_{2} - 1 + . . . . . . + 4 k_{r} - 1 = 4 (k_{1} + k_{2} + . . . . . . . + k_{r}) - r$ .
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 $5050 - 4949 = 101$
$∴ 4 (k_{1} + k_{2} + . . . . . . . . . + k_{r}) - r = 101$
This shows that $r \equiv 3$ .
This leads to $k_{1} + k_{2} + k_{3} = 26$ and one can chose distinct positive integers $k_{1}, k_{2}, k_{3}$ in several ways.

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