HMMT February

Work in $R=\mathbb{Z} / 100 \mathbb{Z} \cong \mathbb{Z} / 4 \mathbb{Z} \times \mathbb{Z} / 25 \mathbb{Z}$. Call an element $r \in R$ type $(s, t)$ if $s=\nu_{2}(r) \leq 2$ and $t=\nu_{5}(r) \leq 2$. Also, define an element $r \in R$ to be coprime if it is of type $(0,0)$, powerful if it is of types $(0,2),(2,0)$, or $(2,2)$, and marginal otherwise. Then, note that if if $r \in R$ is marginal, then any power of $r$ is powerful. Therefore all marginal elements must be in $A$. We claim that all powerful elements are the cube of some marginal element. To show this take a powerful element $r$. In modulo 4 or 25, if $r$ is a unit, then since 3 is coprime to both the sizes of $(\mathbb{Z} / 4 \mathbb{Z})^{\times}$and $(\mathbb{Z} / 25 \mathbb{Z})^{\times}$, it is the cube of some element. Otherwise, if $r$ is zero then it is the cube of 2 or 5, respectively (since this case happens at least once this means that the constructed cube root is marginal). We now claim that 4 additional elements are needed to generate the coprime elements. To see this, note that $R^{\times} \cong \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 20 \mathbb{Z}$ since there are primitive roots $\bmod 4$ and 25. Under this isomorphism, one can show that $(1,1),(1,2),(1,4)$, and $(0,1)$ generate anything, and that no element in $R^{\times}$has more than one of these as a multiple. To wrap up, note that there are $100-(20+1)(2+1)=37$ marginal elements, so 41 elements are needed in total.