Question

Hallar "n" si: A= 28^n*50; tiene 150 divisores

Answer 1

A=(2²·7)^n·(2·5²)
A=2^2n·7^n·2·5²
A=(2^2n+1)·5²·7^n
CD=(2n+1+1)(2+1)(n+1)
150=(2n+2)(3)(n+1)
150=(2n²+4n+2)(3)
150=6n²+12n+6
6n²+12n-144=0
n²+2n-24=0
(n+6)(n-4)=0
n+6=0   ∨  n-4=0
     n=-6        n=4