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Lv 5
Problem Solved asked in Science & MathematicsMathematics · 6 years ago

Prove that the last digit of p^8 is always 1.?

Where p is a prime greater than 5

2 Answers

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  • Amy
    Lv 7
    6 years ago

    Start with the fact that the last digit of p can only be 1, 3, 7, or 9.

    A number ending in 1 or 9, squared, ends in 1.

    A number ending in 3 or 7, squared, ends in 9.

    So p^2 ends in 1 or 9.

    And (p^2)^2 ends in 1.

    And p^8 = ((p^2)^2)^2 ends in 1.

  • ?
    Lv 7
    6 years ago

    p > 5 and is prime

    if the last digit of p = any number other than 1, 3, 7 or 9, then p is not prime.

    last digit = 0, 2, 4, 6 or 8 implies p is divisible by 2.

    last digit = 5 implies p is divisible by 5.

    1^8 -> 1 is last digit

    3^8 -> 1 is last digit

    7^8 -> 1 is last digit

    9^8 -> 1 is last digit

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