Key Concept: Modular Arithmetic and Fermat's Little Theorem

This problem requires us to find the remainder of a large power of an integer when divided by another integer. This falls under the domain of modular arithmetic. A powerful tool for simplifying exponents in modular arithmetic, especially when the divisor is a prime number, is Fermat's Little Theorem.

Fermat's Little Theorem states that if $p$ is a prime number, then for any integer $a$ not divisible by $p$, we have $a^{p-1} \equiv 1 \pmod p$. This theorem allows us to significantly reduce large exponents.

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Step-by-Step Solution

1. Identify the Modulus and Base, and Apply Fermat's Little Theorem

• Understanding the step: We need to find the remainder of $7^{103}$ when divided by 17. Here, the base is $a=7$ and the modulus is $p=17$. Since 17 is a prime number and 7 is not divisible by 17, we can apply Fermat's Little Theorem.
• Applying the theorem: According to Fermat's Little Theorem, $7^{17-1} \equiv 7^{16} \equiv 1 \pmod{17}$. This means that every time we have $7^{16}$, it behaves like 1 in modulo 17 arithmetic.
• Reducing the exponent: Our goal is to reduce the exponent 103 using this property. We divide 103 by 16: $103 = 16 \times 6 + 7$ This tells us that $7^{103}$ can be written as $(7^{16})^6 \times 7^7$.
• Modular reduction: Now we substitute this back into our problem: $7^{103} \equiv (7^{16})^6 \times 7^7 \pmod{17}$ Since $7^{16} \equiv 1 \pmod{17}$: $7^{103} \equiv (1)^6 \times 7^7 \pmod{17}$ $7^{103} \equiv 1 \times 7^7 \pmod{17}$ $7^{103} \equiv 7^7 \pmod{17}$ Explanation: By applying Fermat's Little Theorem, we have reduced the problem of finding the remainder of $7^{103}$ to finding the remainder of $7^7$ when divided by 17, which is a much simpler calculation.

2. Calculate the Remainder of the Reduced Power

• Understanding the step: We now need to find $7^7 \pmod{17}$. We will calculate powers of 7 incrementally, taking the remainder modulo 17 at each step to keep the numbers small.
• Calculations:
  • $7^1 \equiv 7 \pmod{17}$
  • $7^2 = 49$. To find $49 \pmod{17}$: $49 = 2 \times 17 + 15$. So, $49 \equiv 15 \pmod{17}$. Alternatively, $49 \equiv -2 \pmod{17}$ (since $49 = 3 \times 17 - 2$). Using negative remainders can sometimes simplify calculations.
  • $7^3 \equiv 7^2 \times 7 \equiv 15 \times 7 \pmod{17}$ $15 \times 7 = 105$. To find $105 \pmod{17}$: $105 = 6 \times 17 + 3$. So, $105 \equiv 3 \pmod{17}$. (Using negative remainder from $7^2$: $7^3 \equiv (-2) \times 7 \equiv -14 \pmod{17}$. Since $-14 + 17 = 3$, we have $-14 \equiv 3 \pmod{17}$.)
  • $7^4 \equiv 7^3 \times 7 \equiv 3 \times 7 \pmod{17}$ $3 \times 7 = 21$. To find $21 \pmod{17}$: $21 = 1 \times 17 + 4$. So, $21 \equiv 4 \pmod{17}$.
  • $7^5 \equiv 7^4 \times 7 \equiv 4 \times 7 \pmod{17}$ $4 \times 7 = 28$. To find $28 \pmod{17}$: $28 = 1 \times 17 + 11$. So, $28 \equiv 11 \pmod{17}$.
  • $7^6 \equiv 7^5 \times 7 \equiv 11 \times 7 \pmod{17}$ $11 \times 7 = 77$. To find $77 \pmod{17}$: $77 = 4 \times 17 + 9$. So, $77 \equiv 9 \pmod{17}$.
  • $7^7 \equiv 7^6 \times 7 \equiv 9 \times 7 \pmod{17}$ $9 \times 7 = 63$. To find $63 \pmod{17}$: $63 = 3 \times 17 + 12$. So, $63 \equiv 12 \pmod{17}$.

3. Conclusion

From our calculations, we find that $7^{103} \equiv 7^7 \equiv 12 \pmod{17}$.

Therefore, the remainder when $7^{103}$ is divided by 17 is 12.

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Tips and Common Mistakes to Avoid

• Fermat's Little Theorem applicability: Remember that Fermat's Little Theorem applies when the modulus is a prime number and the base is not a multiple of the prime. For composite moduli, Euler's Totient Theorem is used.
• Careful with exponents: Ensure you correctly divide the large exponent by $(p-1)$ and use the remainder. A common mistake is to forget to take the remainder and just use the quotient.
• Modular reduction at each step: Always reduce the result modulo the divisor after each multiplication to keep the numbers manageable. This prevents calculations with very large numbers and reduces the chance of arithmetic errors.
• Using negative remainders: While optional, using negative remainders (e.g., $15 \equiv -2 \pmod{17}$) can often simplify calculations, especially when squaring or cubing numbers. Just remember to convert to a positive remainder at the very end if required. For example, if you end up with $-5 \pmod{17}$, the positive remainder is $-5+17 = 12$.

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Summary and Key Takeaway

To find the remainder of a large power like $a^N \pmod p$ where $p$ is a prime number, the most efficient method is to use Fermat's Little Theorem. First, reduce the exponent $N$ modulo $(p-1)$. Then, calculate the remainder of the base raised to this smaller exponent by performing sequential modular multiplications. This systematic approach ensures accuracy and simplifies complex problems.