MHB Find Last Two Digits of ${2017}^{82}$

[Deanmark]

What are the last two digits of ${2017}^{82}$? We have only learned Fermat's little theorem and Euler's generalization so I am not sure how they apply?

 

[I like Serena]

What are the last two digits of ${2017}^{82}$? We have only learned Fermat's little theorem and Euler's generalization so I am not sure how they apply?

Let's try to apply Euler's generalization.
To find the last two digits we need ${2017}^{82} \bmod 100$, so we have $n=100$.
How far can you get calculating Euler's totient function $\phi(100)$?

 

[Deanmark]

Let's try to apply Euler's generalization.
To find the last two digits we need ${2017}^{82} \bmod 100$, so we have $n=100$.
How far can you get calculating Euler's totient function $\phi(100)$?

I don't know anything about it. We only went of Fermat's and Euler's generalization of ax$\equiv$ b mod(m). The class is a Ring Theory class so we aren't really diving that deeply into the number theory.

 

[MountEvariste]

I don't know anything about it. We only went of Fermat's and Euler's generalization of ax$\equiv$ b mod(m). The class is a Ring Theory class so we aren't really diving that deeply into the number theory.

$\displaystyle \varphi(n) :=n \prod_{p\mid n} \left(1-\frac{1}{p}\right)$ and $100 = 2^25^2$ so $\displaystyle \varphi(100) =100 \prod_{p\mid 100} \left(1-\frac{1}{p}\right) = 100 \left(1-\frac{1}{2}\right) \left(1-\frac{1}{5}\right) = 40. $

Euler's Theorem: if $a$ and $n$ are coprime, then $\displaystyle a^{\varphi(n)} \equiv 1 \bmod{n}$ where $\varphi(n)$ is given by the above product.

Since $2017$ is relatively coprime to $100$, what does that tell you?

 

[Deanmark]

$\displaystyle \varphi(n) :=n \prod_{p\mid n} \left(1-\frac{1}{p}\right)$ and $100 = 2^25^2$ so $\displaystyle \varphi(100) =100 \prod_{p\mid 100} \left(1-\frac{1}{p}\right) = 100 \left(1-\frac{1}{2}\right) \left(1-\frac{1}{5}\right) = 40. $

Euler's Theorem: if $a$ and $n$ are coprime, then $\displaystyle a^{\varphi(n)} \equiv 1 \bmod{n}$ where $\varphi(n)$ is given by the above product.

Since $2017$ is relatively coprime to $100$, what does that tell you?

$2017^{40} \equiv$ 1 mod (100)

$2017^{82} = 2017^{80} \cdot 2017^{2}$

$ 2017^{2} - 1 \equiv 0 \ mod (100)$

88??

 

[MountEvariste]

$ 2017^{2} - 1 \equiv 0 \ mod (100)$

Could you explain what's happening here?

 

[Deanmark]

Could you explain what's happening here?

A mistake.

It's just $2017^{2}$ mod(100) which you can find with a calculator, but if you did not have a calculator could we transform $2017^{2}$ mod(100) further.

 

[MountEvariste]

A mistake.

It's just $2017^{2}$ mod(100) which you can find with a calculator, but if you did not have a calculator could we transform $2017^{2}$ mod(100) further.

We have $2017^{2}\bmod{100} \equiv (17)^2\bmod{100} \equiv 289 \bmod{100} \equiv 89 \bmod{100}. $

By definition if $a=b+km$ for $k \in \mathbb{Z}$ then $\displaystyle a \equiv b \bmod{m}$ (first and last step).

 

[Deanmark]

We have $2017^{2}\bmod{100} \equiv (-17)^2\bmod{100} \equiv 289 \bmod{100} \equiv 89 \bmod{100}. $

By definition if $a=b+km$ for $k \in \mathbb{Z}$ then $\displaystyle a \equiv b \bmod{m}$ (first and last step).

Gotcha. I appreciate it.

 

[kaliprasad]

We have $2017^{2}\bmod{100} \equiv (-17)^2\bmod{100} \equiv 289 \bmod{100} \equiv 89 \bmod{100}. $

By definition if $a=b+km$ for $k \in \mathbb{Z}$ then $\displaystyle a \equiv b \bmod{m}$ (first and last step).

No

We have $2017^{2}\bmod{100} \equiv (17)^2\bmod{100} \equiv 289 \bmod{100} \equiv 89 \bmod{100}. $and further last term 89 shall do

 

[MountEvariste]

No

We have $2017^{2}\bmod{100} \equiv (17)^2\bmod{100} \equiv 289 \bmod{100} \equiv 89 \bmod{100}. $and further last term 89 shall do

Help me understand the objection here? 'Cause $17$ and $-17$ differ by $100$.

Is there some convention you're using? The 'further last term' is obvious.

 

[I like Serena]

Help me understand the objection here? 'Cause $17$ and $-17$ differ by $100$.

Nope. They differ by 34 (or 66 if we calculate the difference the other way round).
We do have $-17 \equiv 100 - 17 \equiv 83 \pmod{100}$.

 

[MountEvariste]

Nope. They differ by 34 (or 66 if we calculate the difference the other way round).
We do have $-17 \equiv 100 - 17 \equiv 83 \pmod{100}$.

Oh, of course. It should be $\mathbf{+}17$. I've edited it now.