If one of the factors of the number $$3^{7} 2^{8} 17^{3}$$ is randomly chosen, then the probability that the chosen factor will be a perfect square is.

The given number is $$3^{7} 2^{8} 17^{3}$$ , lets calculate total no of factors first :

For 3 :$$\left\{3^0,3^1,3^2,......3^7\right\}$$

For 2 :$$\left\{2^0,2^1,2^2,......2^8\right\}$$

For 17 :$$\left\{17^0,17^1,17^2,17^3\right\}$$

So, total no of factors = 8x9x4 = 288

Now, we can find the no of factors which are in the form of perfect squares by considering only powers of even numbers {0,2,4,6,8} because the even power raised can always be written as a perfect square . Hence,

For 3 :$$\left\{3^0,3^2,3^4,3^6\right\}$$

For 2 :$$\left\{2^0,2^2,2^4,2^6,2^8\right\}$$

For 17 :$$\left\{17^0,17^2\right\}$$

Total no of factors in the form of perfect square = 4x5x2 = 40

Therefore the probability of selecting the factor in form of perfect square = $$\frac{\ 40}{288}=\frac{\ 5}{36}$$