π Remainders & Last Digits
The most common number theory pattern in AMC contests: finding remainders and last digits using modular cycles.
π Recognition Cues
Key Phrases:
- “Find the remainder when $a^b$ is divided by $m$”
- “What is the last digit of $a^b$?”
- “Find the remainder when $n!$ is divided by $p$”
- “What is the smallest positive integer $k$ such that $a^k \equiv 1 \pmod{m}$?”
Problem Structure:
- Large exponents with specific moduli
- Questions about final digits
- Cycle or periodicity problems
- Order and primitive root questions
π Solution Template
Step 1: Identify the Pattern
- Last digit: Work modulo 10
- General remainder: Work modulo the given number
- Large exponent: Look for cycles or use Fermat/Euler
Step 2: Find the Cycle
- Check gcd: Ensure $\gcd(a,m) = 1$ for Fermat/Euler
- Compute powers: Find $a^k \equiv 1 \pmod{m}$ for smallest $k$
- Use theorems: Apply Fermat’s Little Theorem or Euler’s theorem
Step 3: Reduce the Exponent
- Modulo cycle length: If $a^k \equiv 1 \pmod{m}$, then $a^b \equiv a^{b \bmod k} \pmod{m}$
- Break down: Use $a^{bc} = (a^b)^c$ for complex exponents
Step 4: Compute the Answer
- Direct calculation: For reduced exponents
- Pattern recognition: Use known cycles for common bases
π― Worked Examples
Example 1: Last Digit Problem
Problem: Find the last digit of $7^{2023}$.
Solution:
- Pattern: Last digit β work modulo 10
- Check gcd: $\gcd(7,10) = 1$ β
- Find cycle: $\varphi(10) = 4$, so $7^4 \equiv 1 \pmod{10}$
- Reduce exponent: $2023 = 4 \cdot 505 + 3$
- Apply: $7^{2023} = 7^{4 \cdot 505 + 3} = (7^4)^{505} \cdot 7^3 \equiv 1^{505} \cdot 7^3 \pmod{10}$
- Compute: $7^3 = 343 \equiv 3 \pmod{10}$
Answer: $3$
Example 2: General Remainder Problem
Problem: Find the remainder when $3^{100}$ is divided by $7$.
Solution:
- Pattern: Remainder β work modulo 7
- Check gcd: $\gcd(3,7) = 1$ β
- Use Fermat: $3^6 \equiv 1 \pmod{7}$ (since $7$ is prime)
- Reduce exponent: $100 = 6 \cdot 16 + 4$
- Apply: $3^{100} = 3^{6 \cdot 16 + 4} = (3^6)^{16} \cdot 3^4 \equiv 1^{16} \cdot 3^4 \pmod{7}$
- Compute: $3^4 = 81 \equiv 4 \pmod{7}$
Answer: $4$
β‘ Quick Techniques
Last Digit Shortcuts
- Powers of 2: Cycle through 2, 4, 8, 6
- Powers of 3: Cycle through 3, 9, 7, 1
- Powers of 7: Cycle through 7, 9, 3, 1
- Powers of 9: Cycle through 9, 1
Common Cycles
- Modulo 4: $a^4 \equiv 1 \pmod{10}$ for $\gcd(a,10) = 1$
- Modulo 3: $a^2 \equiv 1 \pmod{3}$ for $\gcd(a,3) = 1$
- Modulo 5: $a^4 \equiv 1 \pmod{5}$ for $\gcd(a,5) = 1$
Special Cases
- Powers of 0: Always 0
- Powers of 1: Always 1
- Powers of 5: Always end in 5
- Powers of 6: Always end in 6
β οΈ Common Pitfalls
Pitfall: Using Fermat’s Little Theorem when $\gcd(a,p) \neq 1$
- Fix: Check the gcd first, or use case analysis
Pitfall: Forgetting to reduce the exponent modulo the cycle length
- Fix: Always reduce large exponents using the order or totient function
Pitfall: Confusing the cycle length with the modulus
- Fix: The cycle length divides $\varphi(m)$, not necessarily equal to it
π Variants & Extensions
Multi-Step Problems
- Combine multiple modular calculations
- Use Chinese Remainder Theorem for composite moduli
- Apply to polynomial expressions
Order Problems
- Find the smallest positive integer $k$ such that $a^k \equiv 1 \pmod{m}$
- Use the fact that $\operatorname{ord}_m(a) \mid \varphi(m)$
- Test divisors of $\varphi(m)$ in ascending order
Pattern Recognition
- Look for small cycles first
- Use symmetry when possible
- Break down complex expressions
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