Solution: We seek the smallest $n > 0$ such that $2^n \equiv 12 \pmod{25}$.

Solution: We seek the smallest $n > 0$ such that $2^n \equiv 12 \pmod{25}$.

["Title: Solve $ 2^n \equiv 12 \pmod{25} $: Find the Smallest Positive Integer $ n $", "---", "Introduction:\nSolving modular equations can be challenging but rewarding, especially when dealing with powers like $ 2^n \pmod{m} $. In this article, we explore how to efficiently find the smallest positive integer $ n > 0 $ such that:", "$$\n2^n \equiv 12 \pmod{25}\n$$", "This modular equation arises frequently in number theory and cryptography, making it a valuable problem to understand and solve.", "---", "Step 1: Understanding the Problem", "We aim to find the smallest $ n > 0 $ satisfying:", "$$\n2^n \equiv 12 \pmod{25}\n$$", "Since 25 is not prime, but small, we use brute-force computation combined with properties of modular arithmetic to identify the solution.", "---", "Step 2: Compute Powers of 2 Modulo 25", "We compute successive powers of 2 modulo 25 until we find one congruent to 12.", "$$\n\begin{aligned}\n2^1 & = 2 \equiv 2 \pmod{25} \\n2^2 & = 4 \equiv 4 \pmod{25} \\n2^3 & = 8 \equiv 8 \pmod{25} \\n2^4 & = 16 \equiv 16 \pmod{25} \\n2^5 & = 32 \equiv 7 \pmod{25} \\n2^6 & = 64 \equiv 14 \pmod{25} \\n2^7 & = 128 \equiv 3 \pmod{25} \\n2^8 & = 256 \equiv 6 \pmod{25} \\n2^9 & = 512 \equiv 12 \pmod{25} \\n\end{aligned}\n$$", "At $ n = 9 $, we find:", "$$\n2^9 = 512 \equiv 12 \pmod{25}\n$$", "---", "Step 3: Verify Minimality", "We checked all values from $ n = 1 $ to $ n = 8 $, none satisfied $ 2^n \equiv 12 \pmod{25} $. Since $ 2^9 \equiv 12 \pmod{25} $ and no smaller $ n $ works, 9 is the smallest positive solution.", "---", "Step 4: Use of Order and Euler’s Theorem (Advanced Insight)", "Euler’s theorem tells us $ \phi(25) = 20 $, so $ 2^{20} \equiv 1 \pmod{25} $. The powers of 2 modulo 25 repeat with a period dividing 20 (the multiplicative order of 2 modulo 25). This confirms our brute-force search is efficient.", "Moreover, since $ \gcd(2,25)=1 $, the powers of 2 form a cyclic subgroup mod 25, so every residue $ \equiv 12 $ must appear periodically within these 20 steps.", "---", "Conclusion:\nThe smallest positive integer $ n $ such that $ 2^n \equiv 12 \pmod{25} $ is:", "$$\n\boxed{9}\n$$", "This solution demonstrates how modular exponentiation can be tackled systematically—either by direct computation or by leveraging number-theoretic principles.", "---", "Further Reading:\n- Modular exponentiation algorithms\n- Finding discrete logarithms\n- Applications in cryptography and cyclic groups", "---", "Keywords:\n$ 2^n \equiv 12 \pmod{25} $, smallest $ n $, modular arithmetic, power cycle, modular exponentiation, number theory, Cryptography basics", "---", "Meta Description (for SEO):\nDiscover how to find the smallest positive integer $ n $ such that $ 2^n \equiv 12 \pmod{25} $. We show a step-by-step solution using modular exponentiation, verify minimality, and explain underlying principles—ideal for number theory learners and enthusiasts."]

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