If the constant term, in binomial expansion of $$\left(2x^r + \frac{1}{x^2}\right)^{10}$$ is 180, then $$r$$ is equal to ___.

The general term in the expansion of $$\left(2x^r + \frac{1}{x^2}\right)^{10}$$ is given by

$$T_{k+1} = \binom{10}{k}\left(2x^r\right)^{10-k}\left(\frac{1}{x^2}\right)^k = \binom{10}{k} \cdot 2^{10-k} \cdot x^{r(10-k) - 2k}$$

For the constant term, the power of $$x$$ must be zero, so we need $$r(10 - k) - 2k = 0$$.

Solving for $$k$$: $$10r - rk - 2k = 0$$, which gives $$k(r + 2) = 10r$$, so $$k = \frac{10r}{r + 2}$$. $$-(1)$$

Since $$k$$ must be a non-negative integer with $$0 \le k \le 10$$, and the constant term equals 180, we also need $$\binom{10}{k} \cdot 2^{10-k} = 180$$. $$-(2)$$

We test values of $$k$$ in equation $$(2)$$:

For $$k = 8$$: $$\binom{10}{8} \cdot 2^{2} = 45 \times 4 = 180$$. This works.

Substituting $$k = 8$$ into equation $$(1)$$: $$8 = \frac{10r}{r + 2}$$.

Cross-multiplying: $$8(r + 2) = 10r$$, so $$8r + 16 = 10r$$, giving $$2r = 16$$, hence $$r = 8$$.

Therefore, the value of $$r$$ is $$8$$.