Question

WhaT are 2 consecutive odd numbers that have the product of 12,099?

Answer 1

I found:
`109xx111=12099`
or
`-111xx-109=12099`

Explanation:

Consider the condition for your two consecutive odd numbers:
`(2n+1)xx(2n+3)=12,099`
rearranging:
`4n^2+8n-12096=0`
Using the Quadratic Formla you should get:
`n_1=-56`
`n_2=54`
So you have in `(2n+1)xx(2n+3)=12,099`:
`109xx111=12099`
or
`-111xx-109=12099`

Answer 2

I got the same answer as Gio (of course), but set it up a little differently.

Explanation:

When asked to find odd (or even) numbers, it is often a good idea to use `2n+1` for an odd number (and `2n` for an even).

In a problem like this, I like to simply get an equation, find all solution and then pick out the odd (or even) solutions if there are any.

I notice that `12,099` is very close to `12,100`, And also recalling that `121 = 11^2`, so `121 xx 100 = 11^2 xx 10^2`

Of course consecutive odd numbers are exactly `2` away from each other, so

let's call our numbers `x-1` and `x+1`

We want `(x-1)(x+1) = 12,099`

So solve `x^2-1=12,099` (Which we've sort of already done!)

`x^2 = 12,100`

`x= +- 110`

So the numbers are:

`110-1 = 109` and `110+1 = 111` (yes, they are odd)

and

`-110-1 = -111` and `-110+1 = -109` (yes, they're odd too)