Every member of A is in B and at least one member of B is not in A
At least one member of A is not in B
We are given:
\[ A = 36^n - 35^n - 1 = (36^n - 1^n) - 35^n \]
Let’s write it as:
\[ A = 36^n - 1^n - 35^n \]
We use the identity:
\[ a^n - b^n \text{ is divisible by } (a - b) \text{ for all positive integers } n \]
Therefore,
\[ 36^n - 1^n \text{ is divisible by } 36 - 1 = 35 \]
So, the expression \( A = (36^n - 1^n) - 35^n \) is divisible by 35 minus \( 35^n \). But since:
\[ 36^n - 1^n \equiv 0 \pmod{35} \Rightarrow A = 36^n - 1^n - 35^n \equiv -35^n \pmod{35} \]
Now note that \( 35^n \equiv 0 \pmod{35} \Rightarrow A \equiv 0 \pmod{35} \)
Thus, for any integer \( n \), the expression \( A \) is divisible by 35.
Let:
Every value of the expression \( A \) is divisible by 35, hence:
\[ A \subseteq B \]
But not all multiples of 35 (e.g., 35, 70, 105, ...) can be expressed as \( 36^n - 1^n - 35^n \), so:
\[ B \not\subseteq A \Rightarrow A \subset B \]
\[ \boxed{\text{Every member of A is in B, but not every member of B is in A}} \]
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: