A book has some page missing consecutively. The sum of the page number of

missing pages is ${976}$. How many pages are missing there?

## Sylhet - 2014

### Re: Sylhet - 2014

Let's find out the sum of some consecutive numbers greater than 976 with the lowest difference. That is:

$1+2+3+...+44=(41*44)/2=990$

990-976=14 ; the sum of the missing page number.

Look, $14=2+12=2+3+9=2+3+4+5$ ; that we want.

So, there were 4 pages missing.

[Maybe it is the easiest way ]

$1+2+3+...+44=(41*44)/2=990$

990-976=14 ; the sum of the missing page number.

Look, $14=2+12=2+3+9=2+3+4+5$ ; that we want.

So, there were 4 pages missing.

[Maybe it is the easiest way ]

- ahmedittihad
**Posts:**147**Joined:**Mon Mar 28, 2016 6:21 pm

### Re: Sylhet - 2014

You're actually misinterpreting the question.

The problem basically gives us that $(x+1) + (x+2) + ... + (x+y) =976 $. With $x+1$ being the first missing page and $x+y$ the last missing page. So we need to find $y$.

$(x+1) + (x+2) + ... + (x+y) = xy + \dfrac {y(y+1)}{2}=976$

So, $y(x+ \dfrac {y+1}{2})=976$.

Now, $\dfrac {y+1}{2}$ is an integer. So $y$ must be odd. Let's look at the divisors of $976$. They are $1,2,4,8,16,61,122,244,488,976$. The question mentioned several pages so $y \neq 1$ and $y$ is odd and the only odd number except $1$ in the set is $61$. So we get, $y=61$

And there are $61$ pages.

The problem basically gives us that $(x+1) + (x+2) + ... + (x+y) =976 $. With $x+1$ being the first missing page and $x+y$ the last missing page. So we need to find $y$.

$(x+1) + (x+2) + ... + (x+y) = xy + \dfrac {y(y+1)}{2}=976$

So, $y(x+ \dfrac {y+1}{2})=976$.

Now, $\dfrac {y+1}{2}$ is an integer. So $y$ must be odd. Let's look at the divisors of $976$. They are $1,2,4,8,16,61,122,244,488,976$. The question mentioned several pages so $y \neq 1$ and $y$ is odd and the only odd number except $1$ in the set is $61$. So we get, $y=61$

And there are $61$ pages.

Frankly, my dear, I don't give a damn.