What would be the last two digits of the value of $$217^{432}$$?

When finding the last two numbers, we would want to change the number so that it has 1 as its unit digit. 

This can be done if we take $$217^{432}=\left(217^4\right)^{108}$$

Since the 7 raised to power 4 ends in 1, following the cycle of 7, 9, 3, 1,7....

To calculate the value of $$217^4$$, we would only have to consider the value of $$17^4$$, since the 200 would not impact the last two digits. 

17 squared is 289. to find the last two digits of 17^4, we only have to find the last two digits of 89 times 89, which would be 21

So we essentially have a number of the form $$\left(....21\right)^{108}$$

When the number is of the form $$\left(...a1\right)^n$$, the last two digits of the value are $$\left(a\times\ n\right)1$$

Hence, the last two digits of $$\left(...21\right)^{108}$$ would be 61

Therefore, Option B is the correct answer.