Solution: We seek the smallest positive integer $ n $ such that $ n^3 \equiv 888 \pmod{1000} $. This means we want the last three digits of $ n^3 $ to be $888$. We solve this congruence modulo $1000$. Since $1000 = 8 imes 125$, and $8$ and $125$ are coprime, we can use the Chinese Remainder Theorem by solving modulo $8$ and modulo $125$ separately.

["Finding the Smallest Positive Integer $ n $ Such That $ n^3 \equiv 888 \pmod{1000} $: A Step-by-Step Solution Using the Chinese Remainder Theorem", "We are tasked with solving the modular cube congruence:", "$$\nn^3 \equiv 888 \pmod{1000}\n$$", "Since $1000 = 8 \ imes 125$ and $ \gcd(8, 125) = 1 $, the Chinese Remainder Theorem guarantees a unique solution modulo $1000$ if we solve the system:", "$$\n\begin{cases}\nn^3 \equiv 888 \pmod{8} \\nn^3 \equiv 888 \pmod{125}\n\end{cases}\n$$", "We will solve each congruence separately and then combine the results.", "---", "### Step 1: Solve $ n^3 \equiv 888 \pmod{8} $", "First, reduce $888 \mod 8$:", "$$\n888 \div 8 = 111 \quad \Rightarrow \quad 888 \equiv 0 \pmod{8}\n$$", "So we solve:", "$$\nn^3 \equiv 0 \pmod{8}\n$$", "We want $ n $ such that $ n^3 \equiv 0 \pmod{8} $. Testing small integers:", "- $ 0^3 = 0 \equiv 0 $\n- $ 2^3 = 8 \equiv 0 $\n- $ 4^3 = 64 \equiv 0 $\n- $ 6^3 = 216 \equiv 0 \mod 8 $", "In fact, only even $ n $ satisfy $ n^3 \equiv 0 \pmod{8} $, and among even $ n $, $ n \equiv 0 \pmod{2} \Rightarrow n^3 \equiv 0 \pmod{8} $ if $ n \equiv 0 \pmod{2} $? Wait — let's check:", "- $ n \equiv 0 \pmod{2} \Rightarrow n = 2k $, $ n^3 = 8k^3 \equiv 0 \pmod{8} $", "So all even $ n $ satisfy $ n^3 \equiv 0 \pmod{8} $. Therefore:", "$$\nn^3 \equiv 0 \pmod{8} \Rightarrow n \equiv 0 \pmod{2}\n$$", "But we want the smallest solution satisfying both moduli, so we keep this in mind: $ n $ must be even.", "But to be precise, let’s test $ n^3 \equiv 0 \pmod{8} $:", "Try $ n = 0, 2, 4, 6 \mod 8 $:", "- $ 0^3 = 0 \equiv 0 $\n- $ 2^3 = 8 \equiv 0 $\n- $ 4^3 = 64 \equiv 0 $\n- $ 6^3 = 216 \equiv 0 $", "So indeed, all even $ n $ satisfy $ n^3 \equiv 0 \pmod{8} $. So the solution to the first congruence is:", "$$\nn \equiv 0 \pmod{2}\n$$", "But we need the smallest $ n $ such that $ n^3 \equiv 888 \pmod{1000} $, so this gives a starting point: $ n \equiv 0 \pmod{2} $, but we’ll tighten it in the next step.", "---", "### Step 2: Solve $ n^3 \equiv 888 \pmod{125} $", "Now solve:", "$$\nn^3 \equiv 888 \pmod{125}\n$$", "First reduce $888 \mod 125$:", "$$\n125 \ imes 7 = 875 \Rightarrow 888 - 875 = 13 \Rightarrow 888 \equiv 13 \pmod{125}\n$$", "So we solve:", "$$\nn^3 \equiv 13 \pmod{125}\n$$", "We seek the smallest positive $ n $ such that this holds, then combine with modulo 8 via CRT.", "Since $125$ is a power of 5, we solve this using Hensel’s Lemma or trial with modular cube roots.", "We look for $ n \mod 125 $ such that $ n^3 \equiv 13 \pmod{125} $", "Try small values or systematic search.", "Note: We can start by solving modulo 5, then lift.", "Step 2a: Solve $ n^3 \equiv 13 \equiv 3 \pmod{5} $", "Try $ n = 0,1,2,3,4 \mod 5 $:", "- $ 0^3 = 0 $\n- $ 1^3 = 1 $\n- $ 2^3 = 8 \equiv 3 \Rightarrow \ ext{solution } n \equiv 2 \pmod{5} $", "Now lift to $ \mod 25 $, then to $ \mod 125 $", "Step 2b: Lift to $ \mod 25 $", "Let $ n = 2 + 5k $. Plug into $ n^3 \equiv 13 \pmod{25} $", "Compute $ (2 + 5k)^3 = 8 + 3(4)(5k) + 3(2)(25k^2) + 125k^3 = 8 + 60k + 150k^2 + \dots $", "Modulo 25: $ 150k^2 \equiv 0 $, $ 60k \equiv 10k $, $ 8 \equiv 8 $", "So:", "$$\nn^3 \equiv 8 + 60k \equiv 8 + 10k \pmod{25}\n$$", "Set $ 8 + 10k \equiv 13 \pmod{25} \Rightarrow 10k \equiv 5 \pmod{25} $", "Divide both sides by 5: $ 2k \equiv 1 \pmod{5} \Rightarrow k \equiv 3 \pmod{5} $", "So $ k = 3 + 5m $, thus $ n = 2 + 5(3 + 5m) = 17 + 25m $", "So $ n \equiv 17 \pmod{25} $", "Step 2c: Lift to $ \mod 125 $", "Let $ n = 17 + 25m $. Compute $ n^3 \mod 125 $", "$$\nn^3 = (17 + 25m)^3 = 17^3 + 3(17^2)(25m) + 3(17)(25m)^2 + (25m)^3\n$$", "Compute $ 17^3 = 4913 $", "$ 3 \cdot 289 \cdot 25m = 3 \cdot 289 \cdot 25 m = 21675m $\n$ 3 \cdot 17 \cdot 625 m^2 = 31875 m^2 $\n$ 15625 m^3 $", "Now reduce modulo 125:", "Note $ 125 \ imes 197 = 24625 $, so:", "- $ 4913 \mod 125 $: $ 125 \ imes 39 = 4875 $, $ 4913 - 4875 = 38 $\n- $ 21675m \mod 125 $: $ 21675 \div 125 = 173.4 $? Wait: $ 125 \ imes 173 = 21625 $, $ 21675 - 21625 = 50 $ → $ 50m $\n- $ 31875 m^2 \equiv 0 \pmod{125} $ (since $ 125 \mid 31875 $)\n- Higher terms divisible by $ 125 $", "So:", "$$\nn^3 \equiv 38 + 50m \pmod{125}\n$$", "Set equal to 13:", "$$\n38 + 50m \equiv 13 \pmod{125} \Rightarrow 50m \equiv -25 \equiv 100 \pmod{125}\n$$", "So:", "$$\n50m \equiv 100 \pmod{125}\n$$", "Divide equation by $25$: $ 2m \equiv 4 \pmod{5} \Rightarrow m \equiv 2 \pmod{5} $", "So $ m = 2 + 5t $, then $ n = 17 + 25(2 + 5t) = 17 + 50 + 125t = 67 + 125t $", "Thus:", "$$\nn \equiv 67 \pmod{125}\n$$", "So solutions to $ n^3 \equiv 888 \pmod{1000} $ must satisfy:", "$$\n\begin{cases}\nn \equiv 67 \pmod{125} \\nn^3 \equiv 0 \pmod{8} \Rightarrow n \ ext{ even}\n\end{cases}\n$$", "Now find smallest positive $ n $ such that:", "$$\nn \equiv 67 \pmod{125}, \quad n \ ext{ even}\n$$", "Let $ n = 125k + 67 $. We need $ 125k + 67 $ even.", "$125k$ is odd if $k$ odd, even if $k$ even. $67$ is odd.", "So $125k + 67$ is even iff $k$ is odd.", "Smallest $k = 1$: $ n = 125(1) + 67 = 192 $", "Check if $ n = 192 $ satisfies $ n^3 \equiv 888 \pmod{1000} $?", "Compute $ 192^3 $", "First, $ 192^2 = (200 - 8)^2 = 40000 - 3200 + 64 = 36864 $", "Then $ 192^3 = 192 \cdot 36864 $", "Compute $ 192 \cdot 36864 $:", "Break: $ 200 \cdot 36864 = 7,372,800 $, subtract $ 8 \cdot 36864 = 294,912 $", "So $ 7,372,800 - 294,912 = 7,077,888 $", "Now $ 7,077,888 \mod 1000 $ is the last three digits: 888", "Yes!", "So $ 192^3 = 7,077,888 \Rightarrow n^3 \equiv 888 \pmod{1000} $", "And $ n = 192 $ is positive and even, and satisfies the modulo 125 condition.", "Is there a smaller positive solution?", "Check smaller $ k $: $ k = 1 $ gives $ n = 192 $. $ k = 0 \Rightarrow n = 67 $, but $ 67 $ is odd, so fails mod 8 condition.", "Next smaller possibility? $ k = -1 \Rightarrow n = -125 + 67 = -58 $, invalid.", "So $ k = 1 $ gives the smallest valid solution.", "---", "### Final Answer:", "The smallest positive integer $ n $ such that $ n^3 \equiv 888 \pmod{1000} $ is $ \boxed{192} $"]