# 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 $(y^r)'=ry^{r-1}$

If $p\equiv r(mod 4) ; r\in \mathbb{N}^{*} ; 1\leq r\leq 4$ and $\forall{x\in \mathbb{N}} ; 0\leq x\leq 9$ we get:
$x^p\equiv x^r(mod 10)$