For any natural number k , let $$a_{k}=3^{k}$$. The smallest natural number m for which $$\left\{(a_{1})^{1}\times(a_{2})^{2}\times...\times(a_{20})^{20}\right\}<\left\{a_{21}\times a_{22}\times...\times a_{20+m}\right\}$$, is

Given expression is $$\left\{(a_{1})^{1}\times(a_{2})^{2}\times...\times(a_{20})^{20}\right\}<\left\{a_{21}\times a_{22}\times...\times a_{20+m}\right\}$$,

$$\left\{(a_{1})^{1}\times(a_{2})^{2}\times...\times(a_{20})^{20}\right\}$$ = $$\left\{3^1\times3^4\times3^9...\times3^{400}\right\}$$

Sum of square of n natural numbers is $$\frac{n\cdot\left(n+1\right)\cdot\left(2n+1\right)}{6}$$

= $$3^{\dfrac{\left(20\cdot21\cdot41\right)}{6}}$$ = $$3^{2870}$$

On right hand side of inequlaity we have $$\left\{a_{21}\times a_{22}\times...\times a_{20+m}\right\}$$

 = $$3^{21}\times3^{22}\times...\times3^{20+m}$$ = $$3^{21+22+...+20+m}$$

Using the sum of the first (n) natural numbers,

$$1+2+\cdots+n = \frac{n(n+1)}{2}$$

$$21 + 22 + \cdots + (20+m)$$

= $$1+2+\cdots+(20+m) - (1+2+\cdots+20)$$

$$1+2+\cdots+(20+m)=\frac{(20+m)(21+m)}{2}$$

$$1+2+\cdots+20 = \frac{20\cdot21}{2} = 210$$

So, $$21+22+\cdots+(20+m)$$

= $$\frac{(20+m)(21+m)}{2} - 210$$

Expanding, $$(20+m)(21+m)=m^2+41m+420$$

Thus, $$\frac{m^2+41m+420}{2}-210$$

$$= \frac{m^2+41m}{2}$$

Since the bases are equal, we must compare the powers.

$$2870<\frac{m^2+41m}{2} \Rightarrow 5740<m(m+41) $$

Here, we can put in the option to check the minimum value that satisfies the inequality.

56: We get 5740<5264. This is false

57: We get 5740<5586. This is false

58: We get 5740<5742. This is the minimum possible value.