# How space efficient is the Asmuth-Bloom scheme

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)?