If the coefficient of $$x^{10}$$ in the binomial expansion of $$\left(\frac{\sqrt{x}}{5^{1/4}} + \frac{\sqrt{5}}{x^{1/3}}\right)^{60}$$ is $$5^k l$$, where $$l, k \in N$$ and $$l$$ is coprime to $$5$$, then $$k$$ is equal to ______

Find the coefficient of $$x^{10}$$ in $$\left(\frac{\sqrt{x}}{5^{1/4}} + \frac{\sqrt{5}}{x^{1/3}}\right)^{60}$$, expressed as $$5^k \cdot l$$ where $$\gcd(l, 5) = 1$$.

First, we write the general term in the binomial expansion. The $$(r+1)$$th term is given by $$T_{r+1} = \binom{60}{r} \left(\frac{\sqrt{x}}{5^{1/4}}\right)^{60-r} \left(\frac{\sqrt{5}}{x^{1/3}}\right)^r$$. Substituting the powers yields $$= \binom{60}{r} \cdot \frac{x^{(60-r)/2}}{5^{(60-r)/4}} \cdot \frac{5^{r/2}}{x^{r/3}}$$ which simplifies to $$= \binom{60}{r} \cdot 5^{r/2 - (60-r)/4} \cdot x^{(60-r)/2 - r/3}$$.

Next, we determine the value of $$r$$ that makes the power of $$x$$ equal to 10. Setting $$\frac{60-r}{2} - \frac{r}{3} = 10$$ leads to $$\frac{3(60-r) - 2r}{6} = 10$$, then $$180 - 3r - 2r = 60$$, so $$5r = 120$$ and hence $$r = 24$$.

Now, the exponent of 5 in the coefficient is $$\frac{r}{2} - \frac{60-r}{4}$$. Substituting $$r=24$$ gives $$\frac{24}{2} - \frac{36}{4} = 12 - 9 = 3$$.

Therefore, the coefficient has the form $$\binom{60}{24} \cdot 5^3$$, so we need to find the power of 5 dividing $$\binom{60}{24}$$. Using Legendre's formula, the exponent of 5 in $$n!$$ is $$\sum_{i=1}^{\infty} \lfloor n/5^i \rfloor$$. Thus, the power of 5 in $$60!$$ is $$\lfloor 60/5 \rfloor + \lfloor 60/25 \rfloor + \lfloor 60/125 \rfloor = 12 + 2 + 0 = 14$$, in $$24!$$ it is $$\lfloor 24/5 \rfloor + \lfloor 24/25 \rfloor = 4 + 0 = 4$$, and in $$36!$$ it is $$\lfloor 36/5 \rfloor + \lfloor 36/25 \rfloor = 7 + 1 = 8$$. Hence, the power of 5 in $$\binom{60}{24}$$ equals $$14 - 4 - 8 = 2$$.

Adding the exponent from the binomial coefficient to the exponent obtained earlier gives the total power of 5 as $$k = 3 + 2 = 5$$.

The answer is $$\boxed{5}$$.