Find the number of positive integers $$𝑛$$ less than or equal to 100 such that $$𝑛$$ is not divisible by any prime number other than $$2$$ or $$3$$.

A positive integer contains no prime factors other than $$2$$ or $$3$$ precisely when it can be written in the form $$n = 2^{a}\,3^{b}$$ with non-negative integers $$a,b$$.

We must list all such numbers not exceeding $$100$$.

Upper limits for the exponents:
Because $$2^{6}=64 \lt 100 \lt 2^{7}$$, we need only $$0 \le a \le 6$$.
Because $$3^{4}=81 \lt 100 \lt 3^{5}$$, we need only $$0 \le b \le 4$$.

$$n = 2^{a}$$ gives $$1,\,2,\,4,\,8,\,16,\,32,\,64$$ → 7 numbers.

$$n = 3\cdot2^{a} \le 100$$ Allowed $$a=0\text{ to }5$$ (because $$3\cdot2^{5}=96 \le 100$$). Numbers: $$3,\,6,\,12,\,24,\,48,\,96$$ → 6 numbers.

$$n = 9\cdot2^{a} \le 100$$ Allowed $$a=0\text{ to }3$$ (since $$9\cdot2^{4}=144 \gt 100$$). Numbers: $$9,\,18,\,36,\,72$$ → 4 numbers.

$$n = 27\cdot2^{a} \le 100$$ Allowed $$a=0,1$$. Numbers: $$27,\,54$$ → 2 numbers.

$$n = 81\cdot2^{a} \le 100$$ Only $$a=0$$ works. Number: $$81$$ → 1 number.

Adding all cases: $$7 + 6 + 4 + 2 + 1 = 20$$.

Therefore, the count of positive integers $$n \le 100$$ with no prime divisors other than $$2$$ or $$3$$ is $$\mathbf{20}$$.