This answer to another question states that shares in Asmuth-Bloom scheme are $n$ times larger than the secret.

From my understanding of the scheme, the share space is defined by the largest modulo $m_n$ since it defines the largest possible value for a share. So a share would be $m_n/m_0 *2$ times larger than the secret ($*2$ because $m_i$ itself needs to be stored as part of the share too)

For example when the secret space is one byte (256) and $m_n = 337$ it would be $337/256 \approx 2 * 2 = 4$ since the share value itself requires two bytes to be stored, so does the modulo $m_i$.

So my question is: What size does a share in Asmuth-Bloom scheme have (generalized)?


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