Given that the product of n natural numbers is 1995, and the sum of these n natural numbers is also 1995, what is the maximum value of N?

Given that the product of n natural numbers is 1995, and the sum of these n natural numbers is also 1995, what is the maximum value of N?

1995=1×5×3×7×19
N the largest is 1
The number is 1995

Let the function f (x) defined on the natural number set n satisfy f (n) = n + 13 (≤ 2000), f (f (N-18)) (n > 2000), and try to find the value of F (2011) as

f(2011)=f[f(2011-180)]=f[f(1831)]=f(1831+13)=f(1844)=1857

It is known that the function f (n) defined on the set n of natural numbers satisfies f (n + 2) = f (n + 1) - f (n)
It is proved that f (n) is a function with period 6
If f (1) = 1, f (2) = 3, find f (2008)

∵f(n+6)=f[(n+4)+2]=f(n+5)-f(n+4)
=f(n+4)-f(n+3)-f(n+4)
=-f(n+3)
=-[f(n+2)-f(n+1)]
=f(n+1)-f(n+2)
=f(n+1)-[f(n+1)-f(n)]
=f(n)
F (n) is a function with period 6
∵ 2008 divided by 6 quotient 334 remainder 4 ∵ f (2008) = f (4) = f (3) - f (2) = [f (2) - f (1)] - f (2) = - f (1) = - 1