This means \(n^3 \equiv 888 \pmod{8}\) and \(n^3 \equiv 888 \pmod{125}\), since \(1000 = 8 imes 125\) and \(\gcd(8,125)=1\).

["Solving ( n^3 \equiv 888 \pmod{8} ) and ( n^3 \equiv 888 \pmod{125} ): Unlocking Modular Cubic Congruences", "Cryptography, number theory, and modular arithmetic converge in solving problems that often appear in mathematical competitions and cryptanalysis: determining integers ( n ) satisfying certain cubic congruences. This article explores how to solve the system\n[\nn^3 \equiv 888 \pmod{8} \quad \ ext{and} \quad n^3 \equiv 888 \pmod{125},\n]\nnoting that since ( 8 ) and ( 125 = 5^3 ) are coprime (( \gcd(8,125) = 1 )), the Chinese Remainder Theorem (CRT) allows us to combine the solutions into a unique solution modulo ( 1000 = 8 \ imes 125 ).", "---", "### Why Break Down ( n^3 \equiv 888 \pmod{8} ) and ( n^3 \equiv 888 \pmod{125} )?", "We aim to find integers ( n ) such that the cube ( n^3 ) leaves remainder 888 modulo both 8 and 125. By the Chinese Remainder Theorem, since 8 and 125 share no common factors, every simultaneous solution modulo 1000 corresponds to a unique pair ( (n^3 \mod 8, n^3 \mod 125) ). Solving each congruence individually simplifies the overall problem.", "---", "### Step 1: Solve ( n^3 \equiv 888 \pmod{8} )", "First, reduce 888 modulo 8:\n[\n888 \div 8 = 111 \ ext{ exactly} \quad \Rightarrow \quad 888 \equiv 0 \pmod{8}.\n]\nSo,\n[\nn^3 \equiv 0 \pmod{8}.\n]\nWe seek integers ( n ) such that ( n^3 ) is divisible by 8.", "Observation: If ( n ) is even, say ( n = 2k ), then\n[\nn^3 = 8k^3 \equiv 0 \pmod{8}.\n]\nThus, any even ( n ) satisfies ( n^3 \equiv 0 \pmod{8} ).", "Now check odd ( n ):\nOdd integers modulo 8: ( 1, 3, 5, 7 ).\n- ( 1^3 = 1 \equiv 1 \pmod{8} )\n- ( 3^3 = 27 \equiv 3 \pmod{8} )\n- ( 5^3 = 125 \equiv 5 \pmod{8} )\n- ( 7^3 = 343 \equiv 7 \pmod{8} )", "None are divisible by 8. So only even ( n ) work.", "Thus, the congruence\n[\nn^3 \equiv 0 \pmod{8}\n]\nis satisfied if and only if ( n ) is even.", "So,\n[\nn \equiv 0 \pmod{2}.\n]", "This is our first condition.", "---", "### Step 2: Solve ( n^3 \equiv 888 \pmod{125} )", "Now solve\n[\nn^3 \equiv 888 \pmod{125}.\n]\nFirst, reduce 888 modulo 125:\n[\n125 \ imes 7 = 875 \quad \Rightarrow \quad 888 - 875 = 13 \quad \Rightarrow \quad 888 \equiv 13 \pmod{125}.\n]\nSo we seek ( n ) such that\n[\nn^3 \equiv 13 \pmod{125}.\n]", "This is more involved. We proceed via Hensel’s Lemma or trial with modular cube roots, but a smart approach uses brute-force within ( \mathbb{Z}_{125} ) or known algorithms.", "We aim to find an integer ( n ) modulo 125 with ( n^3 \equiv 13 \pmod{125} ).", "Since ( 125 = 5^3 ), and 13 is not divisible by 5 (so cube root likely not divisible by 5), try small values.", "But brute-force over 125 is tedious. Instead, test values congruent modulo 25 first, then lift.", "Alternatively, test ( n ) such that ( n^3 \equiv 13 \pmod{5} ), then lift using Hensel’s Lemma.", "Modulo 5:\n( n^3 \equiv 13 \equiv 3 \pmod{5} )\nTry ( n = 0,1,2,3,4 ):\n- ( 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} ) is the solution.", "Now lift to modulo 25. Let ( n = 2 + 5k ). Then:\n[\nn^3 = (2 + 5k)^3 = 8 + 3(4)(5k) + 3(2)(25k^2) + 125k^3 = 8 + 60k + 150k^2 + 125k^3.\n]\nModulo 25: ( 150k^2 \equiv 0 ), ( 125k^3 \equiv 0 ), so\n[\nn^3 \equiv 8 + 10k \pmod{25} \quad (\ ext{since } 60k \equiv 10k \pmod{25}).\n]\nWe want ( n^3 \equiv 13 \pmod{25} ), so:\n[\n8 + 10k \equiv 13 \pmod{25} \quad \Rightarrow \quad 10k \equiv 5 \pmod{25}.\n]\nDivide equation by 5: ( 2k \equiv 1 \pmod{5} \Rightarrow k \equiv 3 \pmod{5} ).\nSo ( k = 3 + 5m ), and\n[\nn = 2 + 5k = 2 + 5(3 + 5m) = 17 + 25m \quad \Rightarrow \quad n \equiv 17 \pmod{25}.\n]", "Now lift to modulo 125. Let ( n = 17 + 25m ). Compute ( n^3 \mod 125 ):\n[\nn^3 = (17 + 25m)^3 = 17^3 + 3(17^2)(25m) + 3(17)(625m^2) + (25m)^3.\n]\nModulo 125:\n- ( 25^2 = 625 \equiv 0 \pmod{125} ), so terms with ( 625m^2 ) and higher vanish.\nSo:\n[\nn^3 \equiv 17^3 + 3(289)(25m) \pmod{125}.\n]\nFirst, ( 17^3 = 4913 ). ( 4913 \mod 125 ):\n( 125 \ imes 39 = 4875 ), so ( 4913 - 4875 = 38 ).\nThus,\n[\nn^3 \equiv 38 + 3 \ imes 289 \ imes 25m \pmod{125}.\n]\nCompute ( 3 \ imes 289 = 867 ), then ( 867 \ imes 25 = 21675 ). But better:\nModulo 125,\n[\n3 \ imes 289 \ imes 25m = 25m \ imes 867.\n]\nBut ( 867 \mod 5 ): compute ( 867 \div 125 = 6 \ imes 125 = 750 ), ( 867 - 750 = 117 ). So ( 867 \equiv 117 \pmod{125} ), but use modulo ( 125/ \gcd(25,125) = 125/25 = 5 ). Actually, since multiplying by 25,\n[\n3 \cdot 289 \cdot 25m \equiv 25 \cdot (3 \cdot 289 m) = 25 \cdot 867m \pmod{125}.\n]\nNow ( 25 \cdot 867m = 21675m ). But modulo 125:\nNote ( 25 \ imes x \pmod{125} ) depends only on ( x \mod 5 ). Since ( 25 \ imes 5k = 125k \equiv 0 ), only ( x \mod 5 ) matters.", "But instead, compute:\n[\n3 \cdot 289 = 867 \Rightarrow 867 \mod 5 = 2 \Rightarrow 867 \equiv 2 \pmod{5}, \ ext{ but we need mod 5 of } 3 \cdot 289 \cdot 25m.\n]\nBetter:\n[\nn^3 \equiv 38 + 25 \cdot (3 \cdot 289 \cdot m) = 38 + 25 \cdot (867m) \pmod{125}.\n]\nNow ( 25 \cdot 867m = 21675m ). Divide 21675 by 125:\n( 125 \ imes 173 = 21625 ), remainder 50. So\n[\n25 \cdot 867m \equiv 50m \pmod{125}.\n]\nThus,\n[\nn^3 \equiv 38 + 50m \pmod{125}.\n]\nSet equal to 13:\n[\n38 + 50m \equiv 13 \pmod{125} \quad \Rightarrow \quad 50m \equiv -25 \pmod{125} \quad \Rightarrow \quad 50m \equiv 100 \pmod{125} \ ext{ (since } -25 + 125 = 100).\n]\nSo\n[\n50m \equiv 100 \pmod{125}.\n]\nDivide equation by 25:\n[\n2m \equiv 4 \pmod{5} \quad \Rightarrow \quad m \equiv 2 \pmod{5}.\n]\nThus ( m = 2 + 5t ), so\n[\nn = 17 + 25m = 17 + 25(2 + 5t) = 17 + 50 + 125t = 67 + 125t.\n]\nSo,\n[\nn \equiv 67 \pmod{125}.\n]", "---", "### Step 3: Combine Using Chinese Remainder Theorem", "We now solve the system:\n[\nn \equiv 0 \pmod{2}, \quad n \equiv 67 \pmod{125}.\n]\nWe seek ( n \mod 1000 ). So find ( n = 125k + 67 ), and require ( 125k + 67 ) even.\n( 125k ) is odd if"]