Problem - 4160
Let $n$ be any positive integer, show that $$(5n+1)(5n+2)(5n+3)(5n+4)\equiv -1 \pmod{25}$$
$$\begin{array}{rcl} & & (5n+1)(5n+2)(5n+3)(5n+4) \\ &=& ((5n+1)(5n+4))((5n+2)(5n+3)) \\ &=&(25n^2 + 5n + 4)(25n^2+5n + 6)\\ &=&(25n^2+5n)^2 + 10\times(25n^2+5n) + 24 \\ &\equiv & 24 \\ &\equiv&-1\pmod{25} \end{array}$$