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

  • MHB
  • Thread starter Deanmark
  • Start date
In summary, the last two digits of ${2017}^{82}$ are 89. To find the last two digits, we can apply Euler's generalization and calculate the Euler totient function $\varphi(100) = 40$. This tells us that $2017^{40} \equiv 1 \mod 100$. We can then use this to find $2017^{82} \bmod 100$ by breaking it down into $2017^{80} \cdot 2017^{2}$. From there, we can further simplify by finding $2017^{2} \bmod 100$, which is equivalent to $89 \bmod 100$. Therefore, the last two digits of ${
  • #1
Deanmark
16
0
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?
 
Mathematics news on Phys.org
  • #2
Deanmark said:
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 $\phi(100)$?
 
  • #3
I like Serena said:
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.
 
  • #4
Deanmark said:
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?
 
  • #5
June29 said:
$\displaystyle \varphi(n) :=n \prod_{p\mid n} \left(1-\frac{1}{p}\right)$ 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??
 
  • #6
Deanmark said:
$ 2017^{2} - 1 \equiv 0 \ mod (100)$
Could you explain what's happening here?
 
  • #7
June29 said:
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.
 
  • #8
Deanmark said:
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).
 
Last edited:
  • #9
June29 said:
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.
 
  • #10
June29 said:
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
 
  • #11
kaliprasad said:
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.
 
  • #12
June29 said:
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 $-17 \equiv 100 - 17 \equiv 83 \pmod{100}$.
 
  • #13
I like Serena said:
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.
 

FAQ: Find Last Two Digits of ${2017}^{82}$

What is the significance of finding the last two digits of ${2017}^{82}$?

Finding the last two digits of ${2017}^{82}$ can be useful in certain mathematical calculations, such as determining the remainder when dividing large numbers or in modular arithmetic.

What is the general method for finding the last two digits of a number raised to a large power?

The general method is to use the cyclicity of the last two digits of a number when raised to a power. In this case, we can observe that the last two digits of ${2017}^{n}$ repeat in a cycle of 100, where n is any positive integer.

How can we apply the method to find the last two digits of ${2017}^{82}$?

We can rewrite ${2017}^{82}$ as $({2017}^{2})^{41}$. We know that the last two digits of ${2017}^{2}$ are 29, and since 41 is divisible by 100, the last two digits of $({2017}^{2})^{41}$ will also be 29. Therefore, the last two digits of ${2017}^{82}$ are also 29.

Can we use a more efficient method to find the last two digits of ${2017}^{82}$?

Yes, we can use the binomial theorem to simplify the expression and find the last two digits. This method involves finding the remainder when dividing ${2017}^{82}$ by 100 and then taking the last two digits of the remainder. In this case, the remainder is 17, so the last two digits of ${2017}^{82}$ are also 29.

Are there any other applications for finding the last two digits of ${2017}^{82}$?

Yes, in cryptography, the last two digits of a large number raised to a power can be used as a secret key for encryption. Therefore, finding the last two digits of ${2017}^{82}$ can have practical applications in secure communication systems.

Similar threads

Replies
8
Views
2K
Replies
6
Views
3K
Replies
23
Views
2K
Replies
1
Views
2K
Replies
2
Views
1K
Replies
2
Views
1K
Back
Top