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Math Help - Find the rest by division with 7 of the given sum

  1. #1
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    Angry Find the rest by division with 7 of the given sum

    Hey you all,

    I hope you have better mathematical skills than me. I only study math at a low level since i am study to become a teacher. Sadly, this leaves me with only having a basic knowledge and this is probably why I am stuck with this both problems I have been given to solve and my pass and grade depens on it. But i am stuck and becoming desperate, so i hope there is someone who can help me solving it. I don't necessary need to know the whole solution. But maybe some tips and idications how i will solve it. Thank you all so much.

    (by the way I hope i use the right mathematical language, cos i am only used to the german words in the mathematical sense)

    So these are my problems:

    1) Determine the rest of

    \sum\limits_{i=1}^{n} 10^{10^i} when divided by 7.

    2) Prove that


     n^{4k+1} \equiv n\, mod\, m, for all n, k \in \mathbb{N} \iff m|30
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  2. #2
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    Quote Originally Posted by yasste View Post
    Hey you all,

    I hope you have better mathematical skills than me. I only study math at a low level since i am study to become a teacher. Sadly, this leaves me with only having a basic knowledge and this is probably why I am stuck with this both problems I have been given to solve and my pass and grade depens on it. But i am stuck and becoming desperate, so i hope there is someone who can help me solving it. I don't necessary need to know the whole solution. But maybe some tips and idications how i will solve it. Thank you all so much.

    (by the way I hope i use the right mathematical language, cos i am only used to the german words in the mathematical sense)

    So these are my problems:

    1) Determine the rest of

    \sum\limits_{i=1}^{n} 10^{10^i} when divided by 7.

    2) Prove that


     n^{4k+1} \equiv n\, mod\, m, for all n, k \in \mathbb{N} \iff m|30
    Is this for a grade??
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  3. #3
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    yes and there lays the problem of pressure

    hey

    yes i will get a grade for it and that's why I am so under pressure and quite desperate.

    Do you have an idea about it?? And how come you ask cos of the grade?

    See ya,
    Jess
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by yasste View Post
    hey

    yes i will get a grade for it and that's why I am so under pressure and quite desperate.

    Do you have an idea about it?? And how come you ask cos of the grade?

    See ya,
    Jess
    It is a general rule here that we do not help with things that are to be graded.
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  5. #5
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    Even if i do not want anyone to solve it fully?? I mean i only wanna have tips not a full solution. I wanna solve it myself, but i need a start an idea that is missing, but if i go against any rules here, maybe i should leave it to that then
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  6. #6
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    Quote Originally Posted by yasste View Post
    Hey you all,

    I hope you have better mathematical skills than me. I only study math at a low level since i am study to become a teacher. Sadly, this leaves me with only having a basic knowledge and this is probably why I am stuck with this both problems I have been given to solve and my pass and grade depens on it. But i am stuck and becoming desperate, so i hope there is someone who can help me solving it. I don't necessary need to know the whole solution. But maybe some tips and idications how i will solve it. Thank you all so much.

    (by the way I hope i use the right mathematical language, cos i am only used to the german words in the mathematical sense)

    So these are my problems:

    1) Determine the rest of

    \sum\limits_{i=1}^{n} 10^{10^i} when divided by 7.


    Using basic properties of powers and Fermat's Little Theorem, it's easy to prove that 3^{10^i}=4\!\!\!\pmod 7\,\,\,\forall\,i\in\mathbb{N}, so this sum is just 4n\!\!\!\mod 7 ...


    2) Prove that


     n^{4k+1} \equiv n\, mod\, m, for all n, k \in \mathbb{N} \iff m|30

    It should be, of course, 30\mid m ,and you get this at once from n^{4k+1}=n\!\!\!\pmod m\Longleftrightarrow n(n^{4k}-1)=0\!\!\!\pmod m and realizing that n(n^{4k}-1)=n(n^k-1)(n^k+1)(n^{2k}+1) . Check that this expression, for ANY n,k is always a multiple of 30 (i.e., it is always divisible by 2,3, and 5).

    Tonio
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  7. #7
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    Proof

    Hey Tonio,

    thanks very much. You were already a great help since i was able to solve the first and i have determined the rest now.

    With the second, are you sure it must be 30 divides m, because in my text it says if i translate it into english if m is a divider of 30.

    So i only transferred it wrong??

    When i have finished the way you outlined, i still need to do the other way, don't I? I need to show that this congruence is valid if m is a divider of 30, isn't it??

    Thanks for your help, really appreciate it!!
    Jess
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  8. #8
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    Quote Originally Posted by yasste View Post
    Hey Tonio,

    thanks very much. You were already a great help since i was able to solve the first and i have determined the rest now.

    With the second, are you sure it must be 30 divides m, because in my text it says if i translate it into english if m is a divider of 30.

    So i only transferred it wrong??

    When i have finished the way you outlined, i still need to do the other way, don't I? I need to show that this congruence is valid if m is a divider of 30, isn't it??

    Thanks for your help, really appreciate it!!
    Jess

    Yes, I was wrong: it must be m \mid 30. Now, using n=2\,,\,\,k=2 prove that m isn't divided by 2^2\,,\,\,3^2\,\,\,and\,\,\,5^2, and then using n=2\,,\,k=1 prove that no prime different from 2,3, 5 can divide m, so...
    Very nice little exercise!

    Tonio
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  9. #9
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    This argumentation is right then so far??

    To prove is:

    n^{4k+1} \equiv n\ mod\ m \Leftrightarrow m|30

    So for " \Rightarrow"

    I know that

    n^{4k+1} \equiv n\ mod\ m and therefore i can say that
    this is equivalent to
    n(n^k-1)(n^k+1)(n^{2k}+1) \equiv 0\ mod\ m

    This means that m|(n(n^k-1)(n^k+1)(n^{2k}+1)


    And then i use n=2 and k=2
    and i get 2\cdot 3\cdot5\cdot17, right??
    But at this point I can say anymore about m, isn't it?

    But i don't really have an idea how i can prove now that m doesn't divide 2^2 and 3^2 and 5^2. It might be cos i haven't really understood why that is going to help me. Cos if you show that the only dividers are the ones of 30??

    Thank you!
    Last edited by yasste; February 4th 2010 at 10:11 AM. Reason: formulae wrong
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  10. #10
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    Unhappy mh something is wrong

    Hey,

    i have found a mistake in my calculations for the rest if you divide the sum.

    The way i calculated, 3^{10^i} is not congruent to 4\ \text{mod}\ m, because for example is:

    3^{10^2} \equiv 16 \equiv 2\ \text{mod}\ m and not to 4 mod m.

    Can someone please help me??

    Thank you!!
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  11. #11
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    Unhappy mh something is wrong

    Hey,

    i have found a mistake in my calculations for the rest if you divide the sum (exercise 1 of the ones i posted here).

    The way i calculated, 3^{10^i} is not congruent to 4\ \text{mod}\ m, because for example is:

    3^{10^2} \equiv 16 \equiv 2\ \text{mod}\ m and not to 4 mod m.

    Can someone please help me??

    Thank you!!
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