Proof. Note \( 1000 = 2^3 \cdot 5^3. \) With the notation \( \mathbb{Z} / k \mathbb{Z} = \mathbb{Z}_k, \) Corollary 20 in §9.5 of Dummit & Foote says that in such a breakdown, \( \mathbb{Z}_{1000}^\times = \mathbb{Z}_{2^3}^\times \times \mathbb{Z}_{5^3}^\times, \) where \( \mathbb{Z}_k^\times \) is the multiplicative group of units in \( \mathbb{Z}_k \). For any prime \( p, |\mathbb{Z}_{p^\alpha}^\times| = \) #integers less than \( p^\alpha \) relatively prime to \( p^\alpha \) = \( \varphi(p^\alpha) = p^{\alpha-1}(p - 1), \) where \( \varphi \) is the Euler \( \varphi \)-function counting the number of positive integers less than a given positive integer and relatively prime to it.