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The best way to solve this problem is by using a system of equations.
Since the pages are consecutive, one of the equations would be:
x + 1 = y, where x is the page on the left and y is the page on the right.
Since the product of the page numbers equals 2970, the other equation would be x * y = 2970.
So here is your system:
- x + 1 = y
- x * y = 2970
Solve the system by substituting (x + 1) in for y in the second equation.
x * (x + 1) = 2970
x^2 + x = 2970
x^2 + x - 2970 = 0
(x + 54)(x - 54) = 0
Since the page number cannot be negative, x = 54. This means that y = 55. The book is opened to pages 54 and 55.
If you are looking at facing pages, the numbers will be consecutive. A simple way to solve the problem is
n (the number of the first page) x (n+1) (the number of the second page) = 2,970
n^2 + n = 2,970
n^2 + n - 2,970 = 0
Factoring: (n + 55) (n - 54) = 0 (Tip: Find the square root of 2,970, and try the numbers just above and below that number.)
That gives -55 and +54 as your roots. You cannot have a negative page number, so throw out -55, and check +54.
54 x (54 + 1) = 2,970
54 X 55 = 2,970, which is correct. So your answers are pages 54 and 55.
Hope this helps, and good luck!
When you open a book and multiply the two page numbers you see, the page numbers are consecutive integers.
The numbers can be denoted by n and n+1. The product of the two numbers is 2970.
=> n(n+1) = 2970
=> n^2 + n = 2970
=> n^2 + n - 2970 = 0
=> n^2 + 55n - 54n - 2970 = 0
=> n(n + 55) - 54(n + 55) = 0
=> (n - 54)(n + 55) = 0
=> n = 54 and n = -55
We can ignore the negative root.
The page numbers are 54 and 55
well if you open a book, you will find that the two pages in front of you will be consecutive
Given that the two pages are consecutive
let the first page be page X
then the second page is page x+1
By definition of multiplication
well this needs some factoring
first we see there is a 5 in a 2970 (obvious)
i see a two in the number
2+9+7= 18 thats a multiple of 3
297/3=99 which is still a multiple of 3
99/3= 33 which is still a multiple of 3
33/3=11 Youa re left with a prime number
so it is 5,2,3,3,3,11
you need to form two numbers that the difference is one
by calculation, we find out
5*11 = 55
so the factoring is (X+55)(X-54)=0
I dont know a book that has negative page numbers
so the two pages are 54 and 55
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