# Find the rest by division with 7 of the given sum

• Feb 2nd 2010, 09:43 AM
yasste
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 $\in \mathbb{N} \iff m|30$
• Feb 2nd 2010, 02:10 PM
Drexel28
Quote:

Originally Posted by yasste
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 $\in \mathbb{N} \iff m|30$

• Feb 2nd 2010, 02:36 PM
yasste
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.

See ya,
Jess
• Feb 2nd 2010, 02:37 PM
Drexel28
Quote:

Originally Posted by yasste
hey

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

See ya,
Jess

It is a general rule here that we do not help with things that are to be graded.
• Feb 2nd 2010, 02:46 PM
yasste
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
• Feb 2nd 2010, 08:01 PM
tonio
Quote:

Originally Posted by yasste
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
• Feb 3rd 2010, 03:32 PM
yasste
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
• Feb 3rd 2010, 07:48 PM
tonio
Quote:

Originally Posted by yasste
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
• Feb 4th 2010, 10:02 AM
yasste
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!
• Feb 6th 2010, 04:26 PM
yasste
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.

Thank you!!
• Feb 6th 2010, 04:27 PM
yasste
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.