Find the remainder of 54^2010 when divided by 13 using Fermat’s Little Theorem?

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Find the remainder of 54^2010 when divided by 13 using Fermat’s Little Theorem.
Ive tried many different ways and im just not getting the final answer. Help me out please :DDD

✅ Answers

? Best Answer

  • Fermat’s Little Theorem states that:
    a^(p – 1) ≡ 1 (mod p), if a and p are coprimes, and p is prime.

    54 and 13 are coprimes, and 13 is prime, so we can say that:
    54¹² ≡ 1 (mod 13)

    Notice also that:
    2010 = 167 ⋅ 12 + 6.

    So, we have that:
    54²º¹º ≡ (54¹²)^167 ⋅ 54^6 (mod 13)
    54²º¹º ≡ 1^167 ⋅ 54^6 (mod 13)
    54²º¹º ≡ 1 ⋅ 12 (mod 13)
    54²º¹º ≡ 12 (mod 13)

    So, it leaves a remainder of 12. Good luck!

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