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Deanmark
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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?
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?
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 $\phi(100)$?
$\displaystyle \varphi(n) :=n \prod_{p\mid n} \left(1-\frac{1}{p}\right)$ 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. $Deanmark said:I don't know anything about it. We only went of Fermat's and Euler's generalization of ax$\equiv$ b mod(m). The class is a Ring Theory class so we aren't really diving that deeply into the number theory.
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?
Could you explain what's happening here?Deanmark said:$ 2017^{2} - 1 \equiv 0 \ mod (100)$
June29 said:Could you explain what's happening here?
We have $2017^{2}\bmod{100} \equiv (17)^2\bmod{100} \equiv 289 \bmod{100} \equiv 89 \bmod{100}. $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.
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).
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).
Help me understand the objection here? 'Cause $17$ and $-17$ differ by $100$.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
June29 said:Help me understand the objection here? 'Cause $17$ and $-17$ differ by $100$.
Oh, of course. It should be $\mathbf{+}17$. I've edited it now.I like Serena said:Nope. They differ by 34 (or 66 if we calculate the difference the other way round).
We do have $-17 \equiv 100 - 17 \equiv 83 \pmod{100}$.
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.
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.
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.
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.
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.