oxed{i}Question: What is the smallest positive integer whose cube ends in the digits $888$?

["Boxed Solution: What is the Smallest Positive Integer Whose Cube Ends in 888?", "What is the smallest positive integer whose cube ends in exactly 888? This intriguing mathematical quest seeks the value of ( n ) such that:", "[\nn^3 \equiv 888 \pmod{1000}\n]", "In other words, we want the cube of ( n ) to end with the digits 888—meaning ( n^3 \mod 1000 = 888 ).", "---", "### Why This Problem Matters", "Finding an integer whose cube ends in specific digits is more than a curious puzzle—it touches on modular arithmetic and number theory. Solving congruences like ( n^3 \equiv 888 \pmod{1000} ) reveals powerful techniques used in cryptography, computational mathematics, and algorithm design.", "---", "### Step-by-Step Approach", "We aim to solve:\n[\nn^3 \equiv 888 \pmod{1000}\n]", "Since ( 1000 = 8 \ imes 125 ), and 8 and 125 are coprime, we apply the Chinese Remainder Theorem (CRT).", "#### Step 1: Solve modulo 8", "We compute:", "[\nn^3 \equiv 888 \pmod{8}\n]", "Since ( 888 \div 8 = 111 ), we have:", "[\n888 \equiv 0 \pmod{8}\n\Rightarrow n^3 \equiv 0 \pmod{8}\n\Rightarrow n \equiv 0 \pmod{2}\n]", "More precisely, since ( 2^3 = 8 ), any even ( n ) satisfies ( n^3 \equiv 0 \pmod{8} ). So ( n ) must be divisible by 2.", "---", "#### Step 2: Solve modulo 125", "Now solve:", "[\nn^3 \equiv 888 \pmod{125}\n]", "First, reduce 888 mod 125:", "[\n888 \div 125 = 7 \ imes 125 = 875,\quad 888 - 875 = 13\n\Rightarrow n^3 \equiv 13 \pmod{125}\n]", "We now need the cube root of 13 modulo 125.", "This requires finding ( n ) such that ( n^3 \equiv 13 \pmod{125} ) — a nontrivial step due to the modulus being composite and the cube not easily invertible.", "We first solve modulo 5:", "[\nn^3 \equiv 13 \equiv 3 \pmod{5}\n]", "Try ( n = 0,1,2,3,4 \mod 5 ):", "- ( 0^3 = 0 )\n- ( 1^3 = 1 )\n- ( 2^3 = 8 \equiv 3 ) ✅\n- ( 3^3 = 27 \equiv 2 )\n- ( 4^3 = 64 \equiv 4 )", "So ( n \equiv 2 \pmod{5} )", "Now lift using Hensel’s Lemma to solve ( n^3 \equiv 13 \pmod{25} ), then to ( \pmod{125} ).", "---", "#### Step 3: Lift from mod 5 to mod 25", "Let ( n = 2 + 5k ). Plug into ( n^3 \equiv 13 \pmod{25} ):", "Expand ( (2 + 5k)^3 = 8 + 3(4)(5k) + 3(2)(25k^2) + 125k^3 )", "Modulo 25, only first two terms matter:", "[\n(2 + 5k)^3 \equiv 8 + 60k \equiv 8 + 10k \pmod{25}\n]", "Set:", "[\n8 + 10k \equiv 13 \pmod{25} \Rightarrow 10k \equiv 5 \pmod{25}\n]", "Divide equation by 5:", "[\n2k \equiv 1 \pmod{5} \Rightarrow k \equiv 3 \pmod{5}\n]", "So ( k = 3 + 5m ), thus ( n = 2 + 5(3 + 5m) = 17 + 25m \Rightarrow n \equiv 17 \pmod{25} )", "---", "#### Step 4: Lift from mod 25 to mod 125", "Let ( n = 17 + 25m ). Compute ( n^3 \mod 125 ):", "[\nn^3 = (17 + 25m)^3 = 17^3 + 3 \cdot 17^2 \cdot 25m + 3 \cdot 17 \cdot (25m)^2 + (25m)^3\n]", "Modulo 125, terms with ( 25^2 = 625 ) vanish, so only first two terms matter:", "[\n17^3 = 4913\n]\n[\n3 \cdot 289 \cdot 25m = 3 \cdot 289 \cdot 25m = 21675m\n]", "Now modulo 125:", "- ( 4913 \mod 125 ): ( 125 \ imes 39 = 4875 ), ( 4913 - 4875 = 38 )\n- ( 21675m \mod 125 ): ( 21675 \div 125 = 173.4 ), but ( 125 \ imes 173 = 21625 ), so ( 21675 - 21625 = 50 ), so ( 21675m \equiv 50m \pmod{125} )", "So:", "[\nn^3 \equiv 38 + 50m \pmod{125}\n]", "Set equal to 13:", "[\n38 + 50m \equiv 13 \pmod{125} \Rightarrow 50m \equiv -25 \pmod{125} \Rightarrow 50m \equiv 100 \pmod{125}\n]", "(Divide both sides by 25):\n[\n2m \equiv 4 \pmod{5} \Rightarrow m \equiv 2 \pmod{5}\n]", "So ( m = 2 + 5t ), then ( n = 17 + 25(2 + 5t) = 17 + 50 + 125t = 67 + 125t \Rightarrow n \equiv 67 \pmod{125} )", "---", "### Step 5: Combine using CRT", "We now have:", "[\nn \equiv 67 \pmod{125},\quad n \equiv 0 \pmod{2}\n]", "We solve the system:", "Find ( n ) such that:", "[\nn \equiv 67 \pmod{125},\quad n \equiv 0 \pmod{2}\n]", "Check if 67 is even: No. So try ( n = 67 + 125 = 192 )", "Is 192 even? Yes.", "Thus, ( n = 192 ) satisfies:", "- ( 192 \equiv 67 \pmod{125} )\n- ( 192 \equiv 0 \pmod{2} )", "Now verify ( n^3 \equiv 888 \pmod{1000} )", "Compute ( 192^3 ):", "[\n192^2 = 36864\n]\n[\n192^3 = 192 \cdot 36864\n]", "Break it down:", "[\n192 \cdot 36864 = (200 - 8) \cdot 36864 = 200 \cdot 36864 - 8 \cdot 36864\n]\n[\n= 7,372,800 - 294,912 = 7,077,888\n]", "Indeed:\n[\n192^3 = 7,077,888\n]", "The cube ends in 888.", "---", "### Is This the Smallest?", "We derived ( n \equiv 67 \pmod{125} ) and even. The next smaller solution would be ( 67 - 125 = -58 ), not positive. So 192 is the smallest positive integer satisfying the condition.", "---", "## Final Answer", "The smallest positive integer whose cube ends in 888 is:", "[\n\boxed{192}\n]", "This elegant solution illustrates how combining modular arithmetic, Hensel lifting, and the Chinese Remainder Theorem uncovers deep number-theoretic insights—making it both a mathematical triumph and a curiously satisfying puzzle."]