1x2x3x…x35
avem 5..2x5…3x5...4x5…5x5…6x5…7x5
deci 8 de 5 care inmultiti cu nr pare => 8 zerouri
produsul va fi (…….)x 10^8
10^8 divide (.........)x10^8
deci n=8
x=1x2x3x…xn +1
a) pt n=3 => x=1x2x3+1
x=7, e nr prim
b)pt n=4=> x=1x2x3x4+1
x=25, e divizibil cu 5, e si patrat perfect, 25=5^2
c)pt n=5=> x=1x2x3x4x5+1
x=121, 121=11^2