The weight of brilliance

The information of Brilliant.org requires 215 GB of storage on the flash drives, which is

$\begin{aligned} N_\text{bit} &\approx (S_\text{data}=215\text{ GB})\cdot(\gamma_\text{byte}=1.07\times 10^9\text{ bytes per GB})\cdot(\gamma_\text{bit} =8\text{ bits per byte}) \\ &=1.85\times 10^{12}\text{ bits} \end{aligned}$

Moreover, each floating gate transistor requires $N_\text{MOSFET}\approx 10^4$ electrons to register a 1 (each bit is held by one such floating gate transistor), and each electron has the mass $m_e\approx 9.1\times 10^{-31}\text{ kg}$ .

Altogether, this gives a mass of $\hat{m}_\text{Brilliant} = N_\text{bit}N_\text{MOSFET}m_e\approx 1.68\times 10^{-14}\text{ kg}$ .

We are still missing something: the data is information rich. This means that the bits have an equal chance of being in the 0 or 1 state. As an extreme example, the flash drive with all bits in the 1 state contains zero information.

We can show this formally with the entropy of information. The entropy of information is given by

$\begin{aligned} S &\sim -\sum\limits_{i \in \{0,1\}}p_i\log p_i \\ &= -p_0\log p_0 - p_1\log p_1 \end{aligned}$

where $p_0$ and $p_1$ are the probabilities of a given bit being in the 0 or 1 state. This entropy is maximized at $p_0 = p_1 = \frac12$ , i.e. when half of the bits are 1. This means that half the bits are actually in the 0 state, which requires zero electrons to maintain.

Thus, we halve our mass from above and find $\boxed{m_\text{Brilliant} = \hat{m}_\text{Brilliant}/2 = 8.34\times 10^{-15}\text{ kg}}$