Question:** Find the smallest positive integer \(n\) such that \(n^3\) ends in the digits \(888\).

["# Find the Smallest Positive Integer ( n ) Such That ( n^3 ) ends in 888", "Have you ever wondered what the smallest positive integer ( n ) is such that ( n^3 ) ends in the digits 888? In number theory, this problem falls into the fascinating realm of modular arithmetic—specifically, finding a solution to the congruence:", "[\nn^3 \equiv 888 \pmod{1000}\n]", "This task asks you to determine the smallest ( n > 0 ) for which the cube of ( n ) ends in 888. Let’s explore how to solve this step-by-step using smart modular reasoning and computational verification.", "---", "## Understanding the Problem", "We want the last three digits of ( n^3 ) to be 888. Mathematically:", "[\nn^3 \equiv 888 \pmod{1000}\n]", "Because 1000 factors as ( 8 \ imes 125 ), and 8 and 125 are coprime, we can combine results via the Chinese Remainder Theorem (CRT). That is, we solve the system:", "[\n\begin{cases} \nn^3 \equiv 888 \pmod{8} \\nn^3 \equiv 888 \pmod{125}\n\end{cases}\n]", "---", "## Step 1: Solve ( n^3 \equiv 888 \pmod{8} )", "First, reduce 888 modulo 8:", "[\n888 \div 8 = 111 \quad \Rightarrow \quad 888 \equiv 0 \pmod{8}\n]", "So:", "[\nn^3 \equiv 0 \pmod{8}\n]", "We now find all ( n \mod 8 ) such that ( n^3 \equiv 0 \pmod{8} ).", "- Try small values:\n ( n = 0 \Rightarrow 0^3 = 0 \equiv 0 \pmod{8} )\n ( n = 1 \Rightarrow 1 \equiv 1 )\n ( n = 2 \Rightarrow 8 \equiv 0 )\n ( n = 3 \Rightarrow 27 \equiv 3 )\n ( n = 4 \Rightarrow 64 \equiv 0 )\n ( n = 5 \Rightarrow 125 \equiv 5 )\n ( n = 6 \Rightarrow 216 \equiv 0 )\n ( n = 7 \Rightarrow 343 \equiv 7 )", "Thus, ( n^3 \equiv 0 \pmod{8} ) when ( n \equiv 0, 2, 4 \pmod{8} )—all even numbers.", "But let’s be more precise:\n- ( n \equiv 0 \pmod{2} ) implies ( n^3 \equiv 0 \pmod{8} ) only if ( n \equiv 0 \pmod{2} ) and actually ( n ) divisible by 2 gives cubes divisible by 8 when ( n ) is even? Let’s test:", "- Even numbers: ( n = 2k \Rightarrow n^3 = 8k^3 \equiv 0 \pmod{8} )", "Yes! So any even ( n ) satisfies ( n^3 \equiv 0 \pmod{8} ). So:", "[\nn \equiv 0 \pmod{2}\n]", "But we know 888 ≡ 0 mod 8, so condition is: ( n ) must be even.", "So the mod 8 condition is:\n[\nn \equiv 0 \pmod{2}\n]", "---", "## Step 2: Solve ( n^3 \equiv 888 \pmod{125} )", "Now work modulo 125. First reduce 888 mod 125:", "[\n125 \ imes 7 = 875 \quad \Rightarrow \quad 888 - 875 = 13 \quad \Rightarrow \quad 888 \equiv 13 \pmod{125}\n]", "So we must solve:", "[\nn^3 \equiv 13 \pmod{125}\n]", "This is trickier. We seek the smallest positive ( n ) such that ( n^3 \equiv 13 \pmod{125} ).", "Since 125 is ( 5^3 ), we can solve this using Hensel’s Lemma or trial with modular lifting.", "### Sheer computation is feasible since ( \phi(125) = 100 ), so at most 100 residues to check.", "We look for small integers ( n ) such that ( n^3 \equiv 13 \pmod{125} ).", "Try values incrementally:", "- ( n = 1 ): ( 1^3 = 1 )\n- ( n = 2 ): 8\n- ( n = 3 ): 27\n- ( n = 4 ): 64\n- ( n = 5 ): 125 ≡ 0\n- ( n = 6 ): 216 mod 125 = 216 - 125 = 91\n- ( n = 7 ): 343 mod 125: 343 - 2×125 = 343 - 250 = 93\n- ( n = 8 ): 512 mod 125: 512 - 4×125 = 512 - 500 = 12\n- ( n = 9 ): 729 mod 125: 125×5 = 625 → 729 - 625 = 104\n- ( n = 10 ): 1000 ≡ 0\n- ( n = 11 ): 1331 mod 125: 125×10 = 1250 → 1331 - 1250 = 81\n- ( n = 12 ): 1728 mod 125: 125×13 = 1625 → 1728 - 1625 = 103\n- ( n = 13 ): 2197 mod 125: 125×17 = 2125 → 2197 - 2125 = 72\n- ( n = 14 ): 2744 mod 125: 125×21 = 2625 → 2744 - 2625 = 119\n- ( n = 15 ): 3375 ≡ 0\n- ( n = 16 ): 4096 mod 125: 125×32 = 4000 → 4096 - 4000 = 96\n- ( n = 17 ): 4913 mod 125: 125×39 = 4875 → 4913 - 4875 = 38\n- ( n = 18 ): 5832 mod 125: 125×46 = 5750 → 5832 - 5750 = 82\n- ( n = 19 ): 6859 mod 125: 125×54 = 6750 → 6859 - 6750 = 109\n- ( n = 20 ): 8000 ≡ 0\n- ( n = 21 ): 9261 mod 125: 125×74 = 9250 → 9261 - 9250 = 11\n- ( n = 22 ): 10648 mod 125: 125×85 = 10625 → 10648 - 10625 = 23\n- ( n = 23 ): 12167 mod 125: 125×97 = 12125 → 12167 - 12125 = 42\n- ( n = 24 ): 13824 mod 125: 125×110 = 13750 → 13824 - 13750 = 74\n- ( n = 25 ): 15625 ≡ 0\n- ( n = 26 ): 17576 mod 125: 125×140 = 17500 → 76\n- ( n = 27 ): 19683 mod 125: 125×157 = 19625 → 19683 - 19625 = 58\n- ( n = 28 ): 21952 mod 125: 125×175 = 21875 → 21952 - 21875 = 77\n- ( n = 29 ): 24389 mod 125: 125×195 = 24375 → 24389 - 24375 = 14\n- ( n = 30 ): 27000 ≡ 0\n- ( n = 31 ): 29791 mod 125: 125×238 = 29750 → 29791 - 29750 = 41\n- ( n = 32 ): 32768 mod 125: 125×262 = 32750 → 18\n- ( n = 33 ): 35937 mod 125: 125×287 = 35875 → 62\n- ( n = 34 ): 39304 mod 125: 125×314 = 39250 → 54\n- ( n = 35 ): 42875 ≡ 0\n- ( n = 36 ): 46656 mod 125: 125×373 = 46625 → 31\n- ( n = 37 ): 50653 mod 125: 125×405 = 50625 → 28\n- ( n = 38 ): 54872 mod 125: 125×439 = 54875 → too big; 54872 - 54875 = -3 ≡ 122\n- ( n = 39 ): 59319 mod 125: 125×474 = 59250 → 59319 - 59250 = 69\n- ( n = 40 ): 64000 ≡ 0\n- ( n = 41 ): 68921 mod 125: 125×551 = 68875 → 46\n- ( n = 42 ): 74088 mod 125: 125×592 = 74000 → 88\n- ( n = 43 ): 79507 mod 125: 125×636 = 79500 → 7\n- ( n = 44 ): 85184 mod 125: 125×681 = 85125 → 59\n- ( n = 45 ): 91125 ≡ 0\n- ( n = 46 ): 97336 mod 125: 125×778 = 97250 → 86\n- ( n = 47 ): 103823 mod 125: 125×830 = 103750 → 73\n- ( n = 48 ): 110592 mod 125: 125×884 = 110500 → 92\n- ( n = 49 ): 117649 mod 125: 125×941 = 117625 →"]