#1

Can u work this out, we were meant to work it out for our maths homework:

From a book, a number of consecutive pages are missing.

The sum of the page numbers of these pages is 9808.

Which pages are missing?

Thanks to who gets it first xD

From a book, a number of consecutive pages are missing.

The sum of the page numbers of these pages is 9808.

Which pages are missing?

Thanks to who gets it first xD

#2

don't you have the number of pages in the whole book ??

#3

only page 9808 is missing.

or

the pages 291 up to and including 322 are missing.

this is the internet you tart.

or

the pages 291 up to and including 322 are missing.

this is the internet you tart.

#4

There are 32 pages missing starting with page 291

Let X = a page number

Therefore X + 1 = the next consecutive page, x+2, the page after that X+n = the last page and n+1 = total number of pages

If two pages were missing the equation would be

x + (x+1) = 9808

2x + 1 = 9808

2x = 9807

x = 4903.5 which obviously is wrong since you can't be missing half a page( well you can, but you know what I mean)

3 pages would be

x + (x+1) + (x+2) = 9808

3x+3= 9808

4 would be

x + (x+1) + (x+2) + (x+3) = 9808

4x+6

see the pattern of the second nomial? It's a simple arithmetic sequence:

F(n) = Sum (0...N)

and how it relates to the coefficient of the first?

So a general formula can be create thus:

nX + F(n-1) = 9808

nX = 9808 - F(n-1)

X = (9808-F(n-1)) / n

n = number of pages

X = begining page

Let X = a page number

Therefore X + 1 = the next consecutive page, x+2, the page after that X+n = the last page and n+1 = total number of pages

If two pages were missing the equation would be

x + (x+1) = 9808

2x + 1 = 9808

2x = 9807

x = 4903.5 which obviously is wrong since you can't be missing half a page( well you can, but you know what I mean)

3 pages would be

x + (x+1) + (x+2) = 9808

3x+3= 9808

4 would be

x + (x+1) + (x+2) + (x+3) = 9808

4x+6

see the pattern of the second nomial? It's a simple arithmetic sequence:

F(n) = Sum (0...N)

and how it relates to the coefficient of the first?

So a general formula can be create thus:

nX + F(n-1) = 9808

nX = 9808 - F(n-1)

X = (9808-F(n-1)) / n

n = number of pages

X = begining page

#5

By "these pages" you're referring to the missing pages. Either that or the grammar is wrong or I'm wrong :/

#6

hai gaiz. can u do mai home wrkz for me?

#7

Nevermind. I was wrong.

What those guys said up there!

What those guys said up there!

#8

Incomplete.

X.

X.

#9

There are 32 pages missing starting with page 291

Let X = a page number

Therefore X + 1 = the next consecutive page, x+2, the page after that X+n = the last page and n+1 = total number of pages

If two pages were missing the equation would be

x + (x+1) = 9808

2x + 1 = 9808

2x = 9807

x = 4903.5 which obviously is wrong since you can't be missing half a page( well you can, but you know what I mean)

3 pages would be

x + (x+1) + (x+2) = 9808

3x+3= 9808

4 would be

x + (x+1) + (x+2) + (x+3) = 9808

4x+6

see the pattern of the second nomial? It's a simple arithmetic sequence:

F(n) = Sum (0...N)

and how it relates to the coefficient of the first?

So a general formula can be create thus:

nX + F(n-1) = 9808

nX = 9808 - F(n-1)

X = (9808-F(n-1)) / n

n = number of pages

X = begining page

I agree although you way overexplained :-X

It would have been just as simple to say

nx + (n-1) = 9809

#10

I agree although you way overexplained :-X

It would have been just as simple to say

nx + (n-1) = 9809

He pasted that from another website. Not his explanation.

#11

well I don't see why I should bother solving it since only the first one to get it gets your thanks, but probably the easiest way is just (x)+(x+1)+(x+2)=9808 and simplify that to get the first page and just add 1 then 1 again to get the others.

#12

He pasted that from another website. Not his explanation.

This. I take no credit.