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

1. ## 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

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

2) Prove that

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

2. 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

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

2) Prove that

$\displaystyle n^{4k+1} \equiv n\, mod\, m,$ for all n, k $\displaystyle \in \mathbb{N} \iff m|30$
Is this for a grade??

3. ## 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

4. 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.

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.

5. 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

6. 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

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

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

2) Prove that

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

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

Tonio

7. ## 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

8. 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 $\displaystyle m \mid 30$. Now, using $\displaystyle n=2\,,\,\,k=2$ prove that m isn't divided by $\displaystyle 2^2\,,\,\,3^2\,\,\,and\,\,\,5^2$, and then using $\displaystyle n=2\,,\,k=1$ prove that no prime different from 2,3, 5 can divide m, so...
Very nice little exercise!

Tonio

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

To prove is:

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

So for "$\displaystyle \Rightarrow$"

I know that

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

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

And then i use $\displaystyle n=2$ and $\displaystyle k=2$
and i get $\displaystyle 2\cdot 3\cdot5\cdot17$, 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 $\displaystyle 2^2$ and $\displaystyle 3^2$and $\displaystyle 5^2$. 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!

10. ## 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, $\displaystyle 3^{10^i}$ is not congruent to $\displaystyle 4\ \text{mod}\ m$, because for example is:

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

Thank you!!

11. ## 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, $\displaystyle 3^{10^i}$ is not congruent to $\displaystyle 4\ \text{mod}\ m$, because for example is:

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