An epidemiologist analyzing cumulative case data finds a pattern where the number of cases each day follows a cubic growth model. Find the smallest positive integer \( n \) such that \( n^3 \) ends in the digits 001.

["Title: Discovering Hidden Patterns in Disease Spread: How Cubic Growth Models Shape Epidemiological Insights", "Epidemiologists routinely rely on numerical patterns to predict and understand disease transmission. A recent analysis using cumulative case data reveals a surprisingly powerful cubic growth model: daily case counts follow a cubic function ( n^3 ), where ( n ) represents the day number. This cubic trend unveils hidden regularities, enabling researchers to anticipate surges and allocate healthcare resources more effectively.", "But deeper investigation reveals a critical number-theoretic challenge: find the smallest positive integer ( n ) such that ( n^3 ) ends in the digits 001. This seemingly simple constraint unlocks a window into modular arithmetic and Diophantine equations—tools increasingly vital in epidemiological data modeling.", "---", "### Cubic Growth in Epidemiological Data—A Window into Increase Patterns", "In modern outbreak analysis, epidemiologists often model total infections using polynomial functions. While linear models suggest constant daily growth, many pathogens exhibit accelerating spread, especially in early transmission phases—where cubic growth ( n^3 ) emerges naturally. This pattern may reflect compounding effects: each day, the number of new cases increases not just proportionally, but cubically—due to chain reactions, exponential human interactions, or network effects.", "Understanding such growth models allows public health teams to forecast case trajectories with greater precision. But beyond forecasting, uncovering number-theoretic properties—like the smallest ( n ) for which ( n^3 \equiv 1 \pmod{1000} )—can inspire novel computational methods for anomaly detection and pattern recognition in large datasets.", "---", "### Solving the Key Problem: Smallest ( n > 0 ) Such That ( n^3 \equiv 001 \pmod{1000} )", "We seek the smallest positive integer ( n ) for which:", "[\nn^3 \equiv 1 \pmod{1000}\n]", "This means the cube of ( n ) ends in 001. To solve ( n^3 \equiv 1 \pmod{1000} ), we factor 1000 as ( 8 \ imes 125 ), and solve the system:", "[\n\begin{cases}\nn^3 \equiv 1 \pmod{8} \\nn^3 \equiv 1 \pmod{125}\n\end{cases}\n]", "---", "### Step 1: Solve ( n^3 \equiv 1 \pmod{8} )", "Check cubes modulo 8:", "- ( 0^3 = 0 )\n- ( 1^3 = 1 )\n- ( 2^3 = 8 \equiv 0 )\n- ( 3^3 = 27 \equiv 3 )\n- ( 4^3 = 64 \equiv 0 )\n- ( 5^3 = 125 \equiv 5 )\n- ( 6^3 = 216 \equiv 0 )\n- ( 7^3 = 343 \equiv 7 )", "Only ( n \equiv 1 \pmod{8} ) satisfies ( n^3 \equiv 1 \pmod{8} ).", "---", "### Step 2: Solve ( n^3 \equiv 1 \pmod{125} )", "We now solve ( n^3 \equiv 1 \pmod{125} ), where 125 is ( 5^3 ).", "We seek solutions to ( n^3 - 1 \equiv 0 \pmod{125} ), or ( (n - 1)(n^2 + n + 1) \equiv 0 \pmod{125} ).", "So either ( n \equiv 1 \pmod{125} ), or ( n^2 + n + 1 \equiv 0 \pmod{125} ).", "Let’s first test ( n \equiv 1 \pmod{125} ): clearly ( 1^3 = 1 ), so it works.", "But are there others?", "We solve:", "[\nn^2 + n + 1 \equiv 0 \pmod{125}\n]", "Use Hensel’s Lemma to lift solutions from ( \mod 5 ) to ( \mod 125 ).", "Modulo 5:\n( n^2 + n + 1 \equiv 0 \pmod{5} )", "Try ( n = 0,1,2,3,4 ):", "- ( 0 + 0 + 1 = 1 )\n- ( 1 + 1 + 1 = 3 )\n- ( 4 + 2 + 1 = 7 \equiv 2 )\n- ( 9 + 3 + 1 = 13 \equiv 3 )\n- ( 16 + 4 + 1 = 21 \equiv 1 ) → wait, better: ( 2^2 = 4, +2 +1 = 7 \equiv 2 ), ( 3^2 = 9 \equiv 4, +3 +1 = 8 \equiv 3 ), ( 4^2 = 16 \equiv 1, +4 +1 = 6 \equiv 1 )", "None give 0? Wait—check ( n = 2 ): ( 4 + 2 + 1 = 7 \equiv 2 ), ( n = 3 ): ( 9 + 3 + 1 = 13 \equiv 3 ), ( n = 4 ): ( 16 \equiv 1, +4 +1 = 6 ), ( n = 1 ): 3. None ≡ 0.", "But earlier factor: ( n^3 - 1 = (n - 1)(n^2 + n + 1) ). So mod 5, ( n^2 + n + 1 \equiv 0 )? Try ( n = 2 ): ( 4 + 2 + 1 = 7 \equiv 2 ), ( n = 3 ): ( 9 + 3 + 1 = 13 \equiv 3 ), ( n = 4 ): 16+4+1=21≡1, ( n = 0 ): 1. None are 0.", "So only solution mod 5 is ( n \equiv 1 ), and it lifts uniquely?", "Wait—but cubic congruence modulo prime power may have more roots. Let’s verify whether ( n^2 + n + 1 \equiv 0 \pmod{5} ) has solutions.", "Discriminant: ( \Delta = 1 - 4 = -3 \equiv 2 \pmod{5} ). Is 2 a quadratic residue mod 5? Squares: 0,1,4. No. So no solution.", "Thus, only solution mod 5 is ( n \equiv 1 ), and by Hensel’s Lemma (since derivative ( 2n + 1 <br/>\not\equiv 0 \pmod{5} ) for ( n = 1 ): ( 2 + 1 = 3 <br/>\not\equiv 0 )), the lifting to ( \mod 125 ) is unique.", "Therefore, the only solution to ( n^3 \equiv 1 \pmod{125} ) is ( n \equiv 1 \pmod{125} ).", "---", "### Step 3: Combine Using Chinese Remainder Theorem", "We now solve:", "[\n\begin{cases}\nn \equiv 1 \pmod{8} \\nn \equiv 1 \pmod{125}\n\end{cases}\n]", "Since 8 and 125 are coprime, by Chinese Remainder Theorem, ( n \equiv 1 \pmod{1000} ).", "Thus, the smallest positive solution is ( n = 1 )? But ( 1^3 = 1 ), which ends in 001.", "Wait—is that correct?", "Yes! ( 1^3 = 1 ), which ends in 001 when padded to three digits. So the smallest such ( n ) is indeed 1.", "But is there a smaller positive integer? No—1 is the smallest positive integer.", "Wait—could there be another solution?", "Let’s double-check: are there nontrivial cube roots of 1 modulo 1000?", "We concluded only ( n \equiv 1 \pmod{1000} ) satisfies both congruences.", "But let’s test small ( n ) such that ( n^3 \equiv 1 \pmod{1000} ):", "- ( n = 1 ): ( 1^3 = 1 \equiv 1 ) ✅\n- Is there a smaller positive? No.", "But suppose we missed something—could there be a solution like ( n = 501 )? Try:", "( 501^3 = ? )", "But from structure, since ( n \equiv 1 \pmod{8} ) and ( n \equiv 1 \pmod{125} ) → ( n \equiv 1 \pmod{1000} ), only solution is ( n = 1, 1001, \dots )", "But wait—what about ( n = 1 )? Ends in 001? ( 1^3 = 1 ), which is "001" in three digits only with leading zeros—mathematically, ( 1 \equiv 001 \pmod{1000} ), so yes.", "But is there another solution?", "Suppose ( n^3 \equiv 1 \pmod{1000} ), then ( n^3 - 1 = (n - 1)(n^2 + n + 1) \equiv 0 \pmod{1000} )", "Try ( n = 1 ): works.", "Try ( n = 121 ): ( 121^2 = 14641 ), ( +121 + 1 = 14763 ), ( 121 \ imes 14763 )? Too big—check mod 8 and 125.", "But earlier logic shows only solution mod 125 is ( n \equiv 1 ), and mod 8 also requires ( n \equiv 1 ), so CRT gives unique solution ( n \equiv 1 \pmod{1000} )", "But wait—what about ( n \equiv -1 )? ( (-1)^3 = -1 <br/>\not\equiv 1 )", "Or other fractional roots?", "Alternatively, perhaps we made a mistake: are there other cube roots of unity modulo 1000?", "Let’s test ( n = 387 ): too big.", "But known mathematically: the multiplicative order divides ( \phi(1000) = 400 ), and cube roots of 1 form a subgroup of size equal to gcd(3,400)? No—in multiplicative group mod 1000, but 1000 not prime.", "But our modular decomposition is solid.", "Thus, the only solution to ( n^3 \equiv 1 \pmod{1000} ) with ( n > 0 ) and minimal is ( n = 1 ).", "But is ( 1^3 = 1 ) really ending in 001? In decimal representation, 1 is 1, which can be padded to ( \dots001 ) for three digits—yes, in modular contexts, ( n \equiv d \pmod{1000} ) means last three digits are ( d ), so ( d = 1 ) gives 001.", "Therefore, the smallest such ( n ) is:"]