1. ## Euler's Theorem Questions

1) Find the last digit of the decimal expansion of $7^{999,999}$

2) Show that if a is an integer such that a is not divisible by 3 or such that a is divisible by 9, then $a^7$ = a (mod 63)

2. Originally Posted by Janu42
1) Find the last digit of the decimal expansion of $7^{999,999}$

2) Show that if a is an integer such that a is not divisible by 3 or such that a is divisible by 9, then $a^7$ = a (mod 63)
1) "Decimal expansion" is strange terminology to use here. Anyway, gcd(10,7)=1 and eulerphi(10)=4 so....

2) Note that $x\equiv y\pmod{m_1}\land x\equiv y\pmod{m_2}\implies x\equiv y\pmod{[m_1,m_2]}$.

3. Originally Posted by undefined
1) "Decimal expansion" is strange terminology to use here. Anyway, gcd(10,7)=1 and eulerphi(10)=4 so....

2) Note that $x\equiv y\pmod{m_1}\land x\equiv y\pmod{m_2}\implies x\equiv y\pmod{[m_1,m_2]}$.
I don't get what you're doing for #2. I'm confused on the question I think. When it says "such that a is not divisible by 3, or such that a is divisible by 9".... If it was divisible by 9, it's divisible by 3. I don't get what's going on really in this question and how it relates to Euler's.

4. Originally Posted by Janu42
I don't get what you're doing for #2. I'm confused on the question I think. When it says "such that a is not divisible by 3, or such that a is divisible by 9".... If it was divisible by 9, it's divisible by 3. I don't get what's going on really in this question and how it relates to Euler's.
"Or" usually means "inclusive or" but in this case "inclusive or" and "exclusive or" lead to the same interpretation since "not divisible by 3" and "divisible by 9" are mutually exclusive. So we have two cases.

It's fairly standard procedure to break down moduli into prime powers; for example we might do this in order to later apply the Chinese Remainder Theorem.

So let $m_1=9,m_2=7$ and prove that in both cases we get the desired result.

5. Originally Posted by undefined
"Or" usually means "inclusive or" but in this case "inclusive or" and "exclusive or" lead to the same interpretation since "not divisible by 3" and "divisible by 9" are mutually exclusive. So we have two cases.

It's fairly standard procedure to break down moduli into prime powers; for example we might do this in order to later apply the Chinese Remainder Theorem.

So let $m_1=9,m_2=7$ and prove that in both cases we get the desired result.
Let me see if I understand this right. I'm doing two separate cases. For each, am I trying to prove that $a^7$ = x (mod 7) and $a^7$ = y (mod 63), where x is either a or 1 and y is the other? Thereby when I use the CRT I multiply a * 1 to get a making $a^7$ = a (mod 63)?

6. Originally Posted by Janu42
Let me see if I understand this right. I'm doing two separate cases. For each, am I trying to prove that $a^7$ = x (mod 7) and $a^7$ = y (mod 63), where x is either a or 1 and y is the other? Thereby when I use the CRT I multiply a * 1 to get a making $a^7$ = a (mod 63)?
All we want to show is that $a^7\equiv a\pmod{63}$ and to do so we show that $a^7\equiv a\pmod{9}$ and $a^7\equiv a\pmod{7}$.

Case 1: $a\not\equiv0\pmod{3}$

Then gcd(a,9)=1 and $a^7\equiv a\pmod{9}$ by Euler's theorem since eulerphi(9)=6.

Now either $a\equiv0\pmod{7}$ or $a\not\equiv0\pmod{7}$. In the former case, we have $a^7\equiv0^7\equiv0\equiv a\pmod{7}$. In the latter case we have gcd(a,7)=1 and eulerphi(7)=6 so as before $a^7\equiv a\pmod{7}$.

Therefore in this case $a^7\equiv a\pmod{63}$

Case 2: $a\equiv0\pmod{9}$

Try it.

7. Originally Posted by undefined
1) "Decimal expansion" is strange terminology to use here. Anyway, gcd(10,7)=1 and eulerphi(10)=4 so....

2) Note that $x\equiv y\pmod{m_1}\land x\equiv y\pmod{m_2}\implies x\equiv y\pmod{[m_1,m_2]}$.
999,999 is not divisible by any factor of 4 though is it?
I tried doing $(7)^{999999}$ = $((7)^4)^x$ (or some multiple of 4), so that this would be congruent to x (mod 10), where x is the answer I'm looking for.

8. Originally Posted by Janu42
999,999 is not divisible by any factor of 4 though is it?
$999,999 \equiv 3\pmod{4}$

9. Originally Posted by undefined
$999,999 \equiv 3\pmod{4}$
But don't I have to do $(7)^{999999}$ = x (mod10)? Does it follow that x would be 3?

10. Originally Posted by Janu42
But don't I have to do $(7)^{999999}$ = x (mod10)? Does it follow that x would be 3?
$7^{999999}\equiv7^{4\cdot249999+3}\equiv7^{4\cdot2 49999}\cdot7^3\equiv(7^4)^{249999}\cdot7^3\equiv1^ {249999}\cdot7^3\equiv7^3\pmod{10}$

But when you see how it works you skip all the intermediate steps

$7^{999999}\equiv7^{999999\ \text{'mod'}\ \varphi(10)}\pmod{10}$

So now all you need to do is find out what the last digit of $\,7^3$ is.