Problem - 3932
Let real numbers $a$ and $b$ satisfy $0 < a < a +\frac{1}{2} \le b$ and $a^{40}+b^{40}=1$. Show that all the twelve digits after the decimal point are $9$ if $b$ is expressed in decimal.
From the given conditions, we must have $b < 1$ and $a < \frac{1}{2}$. Now, because
$$2^{40} = 1024^4 > 1000^4 = 10^{12} \implies a^{40} < \frac{1}{2^{40}} < \frac{1}{10^{12}}$$
It follows that $$1 > b > b^{40} = 1-a^{40} > 1 - \frac{1}{10^{12}}$$
This means the decimal expression of $b$ must have at least twelve $9$ after the decimal point.