1. Write the general term

Let $a=\frac{\sqrt{x}}{5^{1/4}}=x^{1/2}5^{-1/4},\qquad b=\frac{\sqrt{5}}{x^{1/3}}=5^{1/2}x^{-1/3}.$

Then $(a+b)^{60}$ has general term $T_{r+1}=\binom{60}{r}a^{60-r}b^r.$

So, $T_{r+1}=\binom{60}{r}\left(x^{1/2}5^{-1/4}\right)^{60-r}\left(5^{1/2}x^{-1/3}\right)^r.$

2. Find the power of $x$

Power of $x$ in $T_{r+1}$ is $\frac{60-r}{2}-\frac{r}{3}.$

We need coefficient of $x^{10}$, so $\frac{60-r}{2}-\frac{r}{3}=10.$

Multiply by $6$: $3(60-r)-2r=60$ $180-3r-2r=60$ $180-5r=60$ $5r=120$ $r=24.$

3. Compute the corresponding coefficient

Now substitute $r=24$: $\text{coefficient}=\binom{60}{24}\left(5^{-1/4}\right)^{36}\left(5^{1/2}\right)^{24}.$

Since $60-r=36$, $\left(5^{-1/4}\right)^{36}=5^{-9},\qquad \left(5^{1/2}\right)^{24}=5^{12}.$

Hence, $\text{coefficient}=\binom{60}{24}5^{-9+12}=\binom{60}{24}5^3.$

So we need the exponent of $5$ in $\binom{60}{24}$.

4. Find power of 5 in $\binom{60}{24}$

Using Legendre’s formula: $v_5(60!)=\left\lfloor\frac{60}{5}\right\rfloor+\left\lfloor\frac{60}{25}\right\rfloor=12+2=14,$ $v_5(24!)=\left\lfloor\frac{24}{5}\right\rfloor=4,$ $v_5(36!)=\left\lfloor\frac{36}{5}\right\rfloor+\left\lfloor\frac{36}{25}\right\rfloor=7+1=8.$

Therefore, $v_5\left(\binom{60}{24}\right)=14-4-8=2.$

Thus, $\binom{60}{24}=5^2\cdot l,$ where $l$ is co-prime to $5$.

So the total coefficient is $5^3\cdot 5^2\cdot l=5^5\cdot l.$

Hence, $k=5.$

5. Comparison with stored answer

Stored correct answer = $5$.

This matches our derived answer.