On the board is written the number $18$. Every minute the number is replaced by the product with 2 or 3,or by the quotient of the division with 2 or 3. Show that after $60$ minutes the number cannot be $96$.
What I tried is to factorize into prime: $18=2\cdot3^2$ and $96=2^5\cdot3$, but I don't see what the invariant is, since the division by $2$ or $3$ may result in new prime factors.
On
Denote by $\Omega(n)$ the number of prime factors of $n$ counting multiplicity, so if $n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_r^{\alpha_r}$ then $\Omega(n)=\alpha_1+\alpha_2+\dots + \alpha_r$.
Initially the number on the board $N$ is $18=2\cdot3^2$ so $\Omega(N)$ is odd. After every step we add $1$ or substract $1$ from one of the prime exponents, and thus $\Omega(N)$ changes parity at each operation.
After $60$ minutes the parity of $\Omega(N)$ has changed $60$ times, so it has stayed the same, in other words it is odd.
On the other hand $\Omega(96)=\Omega(2^53^1)=6$ which is even.
Incidentally $\Omega(N)$ is a very studied function in analytic number theory, it even has an OEIS page
Hint: At any time, the number has the shape $2^a\cdot 3^b$. A useful invariant is the parity of $a+b$, that is, the remainder when $a+b$ is divided by $2$.
The parity changes every minute, so it must be unchanged after an even number of minutes.