1.
Obviously, n≥7.
235236n=2n5+3n4+5n3+2n2+3n+6
This holds true for any n≥7.
Then, clearly, 2n5+3n4+5n3+2n2+3n≡1(mod7).
Define P(n)=2n5+3n4+5n3+2n2+3n−1.
Consider all possible values of P(n) mod 7 when 0≤n≤6,n∈Z.
P(0)≡6(mod7)P(1)≡0(mod7)P(2)≡4(mod7)P(3)≡1(mod7)P(4)≡1(mod7)P(5)≡1(mod7)P(6)≡1(mod7)
Generally, P(7k+1)≡0(mod7)∀k∈Z+.
All possible values of n are: 8,15,22,29,36,43,50,57,64,71,78,85,92,99.
So there are 14 possible values of n such that 235236n is divisible by 7.