A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is ______ (Nearest integer)
(Given: antilog 0.125 = 1.333, antilog 0.693 = 4.93)

We need to find the percentage of original activity remaining after 83 days for a radioactive element with a half-life of 200 days. Half-life (t$$_{1/2}$$) = 200 days, time elapsed (t) = 83 days, antilog 0.125 = 1.333, antilog 0.693 = 4.93.

The decay constant is given by $$\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{200}$$. Using the radioactive decay formula $$\frac{A}{A_0} = e^{-\lambda t}$$ and taking log$$_{10}$$ of the ratio, we get $$\log_{10}\left(\frac{A_0}{A}\right) = \frac{\lambda t}{2.303} = \frac{0.693 \times 83}{200 \times 2.303}$$ which yields $$\log_{10}\left(\frac{A_0}{A}\right) = \frac{57.519}{460.6} = 0.1249 \approx 0.125$$.

Using the given antilog value, $$\frac{A_0}{A} = \text{antilog}(0.125) = 1.333$$ so $$\frac{A}{A_0} = \frac{1}{1.333} = 0.750$$ and hence $$\% \text{ activity remaining} = 0.750 \times 100 = 75\%$$. Hence, the answer is 75.