What is x in this arithmetic problem?

[BobG]

Easy arithmetic problem:

x=222,222,222,222,222,222,222^2-222,222,222,222,222,222,221^2

Find x.

 

[kevinferreira]

<br /> x=(n+1)^2-n^2=2n+1<br />
where n=222,222,222,222,222,222,221.
Therefore
<br /> x=444,444,444,444,444,444,443<br />

 

[regor60]

I set N = 222,222,222,222,222,222 so that question becomes

(n+222)^2 - (n+221)^2

after which solving for n is straightforward

Not as nice as the first response, however

 

[Tommaso_Russo]

x=(n+1)^2-n^2=2n+1
where n=222,222,222,222,222,222,221.
Therefore
x=444,444,444,444,444,444,443

mmm...

x=a^2-b^2=(a+b)(a-b)
but a-b = 1 so
x = a+b = 444,444,444,444,444,444,443