INEQUALITY'S BLOG

October 17, 2012

Last two digit (vankhea formula)

Filed under: the new math — KKKVVV @ 11:25 am

If p\equiv r(mod 20); r\in \mathbb{N}^{*}; 1\leq r\leq 20 and \forall{x, y\in \mathbb{N}}; 0\leq x\leq 9; 0\leq y\leq 9. Then we get:
(\overline{xy})^p\equiv y^p+10x(y^r)'(mod 100) with (y^r)'=ry^{r-1}

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a comment