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Math Help - 2 Modulo questions

  1. #1
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    2 Modulo questions

    We're working with modulos in my number theory class, so I'm sure that's how I am supposed to go about doing these problems. I just can't think of the proper approach...

    a)Find the remainder when 5^10 is divided by 19.

    b)Find the final digit of 1! + 2! + 3! + ... + 10!
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  2. #2
    Super Member Bacterius's Avatar
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    These questions usually call for a divide and conquer approach. That is, simplify the problem by breaking it into bits.
    a)Find the remainder when 5^10 is divided by 19.
    Note that 5^{10} = (5^2)^5, and that 5^2 \equiv 25 \equiv 6 \pmod{19}. Thus (5^2)^5 \equiv 6^5 \pmod{19}. Note that 6^5 = 6^2 \times 6^2 \times 6^1, and 6^2 \equiv 36 \equiv 17 \pmod{19}. So 6^5 \equiv 17 \times 17 \times 6 \pmod{19}. Note that 17^2 \equiv 289 \equiv 4 \pmod{19}, so 17^2 \times 6 \equiv 4 \times 6 \equiv 24 \equiv 5 \pmod{19}. So 5^{10} \equiv 5 \pmod{19}.

    b)Find the final digit of 1! + 2! + 3! + ... + 10!
    To find the last digit of this expression, consider it modulo 10 (for example, the last digit of 718 is 718 \equiv 8 \pmod{10}). Let's start with the small factorials : 1! = 1, 2! = 2, 3! = 6. Now 4! \equiv 3! \times 4 \equiv 6 \times 4 \equiv 24 \equiv 4 \pmod{10}. Next one : 5! \equiv 4! \times 5 \equiv 4 \times 5 \equiv 20 \equiv 0 \pmod{10}. Continue like this up to 10! (note that 10! \equiv 0 \pmod{10} since it divides 10) ,then sum up all the last digits and take the last digit of this sum and you are done

    Does it make sense ?
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  3. #3
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    5! up to 10! will all be congruent to 0 (mod 10) right?

    5! = 4! x 5 = 4 x 5 = 20 = 0 (mod 10)
    6! = 5! x 6 = 0 x 6 = (0 mod 10)
    7! = 6! x 7 = 0 x 7 = (0 mod 10)
    ...and so on

    So 1+2+6+4 = 13 which means my answer is 3?
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  4. #4
    Super Member Bacterius's Avatar
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    Yes indeed you are right, since 5! divides 2 and 5 so it divides 10. And so the last digit of this sum of factorials is indeed 3
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  5. #5
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    Hello, jzellt!

    You are right!


    b)Find the final digit of S \;=\;1! + 2! + 3! + 4! + \hdots + 10!

    We have: . S \;=\;1 + 2 + 6 + 24 + 5! + 6! + \hdots + 10!

    \text{Hence: }\;S \;=\;33 + \underbrace{5! + 6! + \hdots + 10!}_{\text{All of these end in 0}}
    . .
    All the factorials, starting with 5!, contain a factor of 10 (at least).


    Therefore, S ends in 3.

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