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]

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

 

[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
$$x = a+b = 444,444,444,444,444,444,443$$