Question

In a book with page numbers from 1 to 100, some pages are torn off. The sum of total numbers on the remaining pages is 4949. How many pages are torn off?

Answer 1

Hint: In this problem we will subtract the sum of remaining pages from the total sum of pages and then we will calculate the number of pages by using the formula of first n integers.
Complete step by step answer:
Given: We have given the total number of pages which are 1 to 100.
And the sum of the number of pages remaining untorn is 4949.
We have to find the number of pages which are torn.
We know that the formula for the sum of first of (n) integer numbers is $\dfrac{{n\left( {n + 1} \right)}}{2}$ .
We have pages 1 to 100.
So, the value of $n$ is 100.
After putting a value of $n$ we will get.
$
   = \dfrac{{100\left( {100 + 1} \right)}}{2} \\
   = 5050 \\
$
So, the sum of the total pages is 5050.
Now, we will find the sum of numbers of torn pages by subtracting the sum of untorn pages from total sum.
Mathematically, $ = 5050 - 4949 = 101$
So, the pages that have been torn off must total to $101$ .
It can be page 1 and page 100.
Page 11, page 40 and page 50.
Etc.
The most number of pages ripped out is 13 because the smallest total that can be achieved with 14 pages is the first 14 pages which totals 105.
But, when we look for consecutive pages then notice that any sequence of $n$ consecutive pages will be the first $n$ pages shifted of an integer multiple of $n$.
So, we required,$\dfrac{{n\left( {n + 1} \right)}}{2} + an = 101$ has an integer solution for $a$ .
This happen when greatest common divisor of \[\left( {n,\dfrac{{\left( {n + 1} \right)}}{2}} \right),\left( {n,\dfrac{{n\left( {n + 1} \right)}}{2}} \right)\] divides 101.
The greatest common divisor for $\left( {n,\dfrac{{n\left( {n + 1} \right)}}{2}} \right)$is $n$, when $n$ is odd and $\dfrac{n}{2}$ ,when $n$ is even.
Because 101 is the prime number, its divisor is 1. Which we get when $n = 2$ .
Therefore, the only consecutive page solution is two pages, page 50 and page 51.

Note: In this problem we have lots of solutions of the number of pages ripped out could be any number from 2 to 13. But if we need consecutive pages then only 2 pages are there.