Determination of the last three digits of $2014^{2014}$

Question

May I know if my proof is correct?

Consider $x$ such that $2014^{2014} \equiv x \pmod{1000}.$

By Euler's theorem, $2014^{\ \psi(1000)} =2014^{\ \psi(2^3)\psi(5^3)}=2014^{400}\equiv 1 \pmod{1000}$.

It follows that $2014^{2000} \equiv 1 \pmod{1000},$ hence $2014^{\ 14} \equiv x \pmod{1000}.$

By Binomial expansion, $2014^{\ 14} = 14^{\ 14}+2000m$, for some positive integer $m$.

Finally$, 14^{14} \equiv x \pmod{1000} \Longleftrightarrow \ x= 16$ and the last three digits are $016.$

In general, could we use the above method to determine the exact value of $n^n, \forall n \in \mathbb{N} ?$

Thank you and wish you a successful 2014!

$n^{\varphi(x)} \equiv 1 \pmod x$ if and only if $ \gcd(x,n)=1$ — Steven Alexis Gregory, Oct 20 '17 at 04:11

score 2 · Answer 1 · answered Dec 31 '13 at 16:12

2

Your proof is not correct because you have to treat the moduli $8$ and $125$ separately (because $8$ and $2014$ are not relatively prime), and use (a very simple case of) the Chinese remainder theorem to deduce the answer. If you do so, you will see that it does not change your result ($16$). Also, you should write $\varphi$ instead of $\psi$.

answered Dec 31 '13 at 16:12

Phira

21,206

Thank you for the advice. I thought if $a,b$ are co-prime, then $\varphi(ab)=\varphi(a)\varphi(b)?$ – Alexy Vincenzo Dec 31 '13 at 16:22
1

@AlexyVincenzo It is true that it is multiplicative, but the Euler-Fermat theorem that you used is only true if the base is coprime to the modulus. – Phira Dec 31 '13 at 16:27

lab bhattacharjee · Accepted Answer · 2014-01-02T11:07:43.433

2

Carmichael function, Euler's totient theorem and Fermat's little theorem need $(a,m)=1$ for $a^n\equiv1\pmod m$ where $n$ may have different values in each case.

But $(2014,1000)=2>1$

As $\displaystyle1000=8\cdot125$

Clearly, $\displaystyle2014^{2014}\equiv0\pmod{2^3}\ \ \ \ (1)$

As $2014\equiv14\pmod{125},\displaystyle 2014^{2014}\equiv14^{2014}\pmod{125}$

As $\phi(125)=100, 2014\equiv14\pmod{100};\displaystyle 14^{2014}\equiv14^{14}\pmod{125}$

Now, $14^{14}=2^{14}\cdot7^{14}$

and $\displaystyle2^7=128\equiv3\pmod{125}\implies 2^{14}\equiv3^2\equiv9$

$\displaystyle7^{14}=(7^2)^7=(50-1)^7\equiv-1+\binom7150\pmod{125}\equiv99$

$\displaystyle\implies 14^{14}\equiv99\cdot9\pmod{125}\equiv16\ \ \ \ (2)$

Now apply the famous CRT on $(1),(2)$

edited Jan 02 '14 at 11:07

answered Dec 31 '13 at 16:16

lab bhattacharjee

279,016

Thank you. May I know why $2014^{2014} \equiv 14^{14} \ (\text{mod} 125)?$ How does it follow from my original argument? – Alexy Vincenzo Dec 31 '13 at 17:47
@AlexyVincenzo, you have found $$2014^{2014}\equiv14^{14}\pmod{2000}$$ As $125$ divides $2000,$ $$2014^{2014}\equiv14^{14}\pmod{125}$$ – lab bhattacharjee Dec 31 '13 at 18:04
I think I have found $2014^{14} \equiv 14^{14} \ (\text{mod 2000})$ instead. – Alexy Vincenzo Dec 31 '13 at 18:08
@AlexyVincenzo, same logic can be applied as $125$ divides $2000$ as well – lab bhattacharjee Dec 31 '13 at 18:09
Sorry, but I still don't get how it follows that $2014^{2014} \equiv 14^{14} \ (\text{mod} \ 2000).$ – Alexy Vincenzo Dec 31 '13 at 18:19
@AlexyVincenzo, You have written "$2014^{\ 14} = 14^{\ 14}+2000m$" right? Now see http://mathworld.wolfram.com/Congruence.html – lab bhattacharjee Dec 31 '13 at 18:21
Thank you but I'm afraid it still doesn't answer my question. – Alexy Vincenzo Dec 31 '13 at 18:24
let us continue this discussion in chat – lab bhattacharjee Dec 31 '13 at 18:25

score 2 · Answer 3 · answered Dec 31 '13 at 16:30

2

You cannot directly apply Euler's theorem because $\,2104\,$ is not coprime to your modulus $\,1000.\,$ But your argument is quickly patched. It shows that $\, n := 2014^{2014}\!\equiv 16\pmod {125}$ and, since $\,8\mid n,\,$ also $\,n\equiv 16\pmod 8\,$ thus $\,n \equiv 16\,$ mod lcm $\!(8,25) = 8\cdot 125 = 1000.$

answered Dec 31 '13 at 16:30

Bill Dubuque

282,220

Thank you. How we know $2014^{2014} \equiv 16 \ (\text{mod} \ 125)?$ – Alexy Vincenzo Dec 31 '13 at 17:03
@Alexy Euler's Phi Theorem applies here, so you can proceed as you did in your question. – Bill Dubuque Dec 31 '13 at 17:06
Using that theorem, I obtain $2014^{200} \equiv 1 \ (\text{mod} 125)$ and hence $2014^{2014} \equiv 1 \ (\text{mod} 125)$ – Alexy Vincenzo Dec 31 '13 at 17:12
But $,14^{14}!\equiv 16\pmod {125},$ so you have an error somewhere. – Bill Dubuque Dec 31 '13 at 17:15
I see. By Euler phi theorem, we get $2014^{100} \equiv 1 \ (\text{mod} \ 125)$ – Alexy Vincenzo Dec 31 '13 at 17:20
@Alexy Right, you original approach works fine modulo $125,,$ using $,\phi(125) = 100$. – Bill Dubuque Dec 31 '13 at 17:21
But how do I get $2014^{2014} \equiv 16 \ (\text{mod 125})$ from here? – Alexy Vincenzo Dec 31 '13 at 17:23
@Alexy One way to finish is to calculate $,14^{14}\pmod{125},,$ e.g. by repeated squaring. – Bill Dubuque Dec 31 '13 at 17:25
Sorry Sir, I don't get you. – Alexy Vincenzo Dec 31 '13 at 17:31
@Alexy Where, precisely are you stuck? – Bill Dubuque Dec 31 '13 at 17:36
What you mean by 'repeated squaring'? – Alexy Vincenzo Dec 31 '13 at 17:43
See e.g. this Wikipedia article. A search on MSE will turn up many examples. – Bill Dubuque Dec 31 '13 at 17:44

score 1 · Answer 4 · answered Dec 31 '13 at 18:15

First of all, the last $n$ digits necessarily implies $\pmod{10^n}$ where $n$ is a positive integer

You have already found $$2014^{2014}\equiv14^{14}\pmod{2000}\implies2014^{2014}\equiv14^{14}\pmod{1000}$$

Now as $\displaystyle14^{14}=2^{14}\cdot7^{14}$ and $(14^{14},1000)=8=2^3$

Let us find $\displaystyle2^{14-3}\cdot7^{14}\pmod{\frac{1000}{2^3}}\implies 2^{11}\cdot7^{14}\pmod{125}$

From my other answer, $\displaystyle7^{14}\equiv99\pmod{125}$

and $\displaystyle2^7=128\equiv3\pmod{125}\implies2^{11}\equiv2^4\cdot3\equiv48\pmod{125}$

$\displaystyle\implies 2^{11}\cdot7^{14}\equiv48\cdot99$ $\displaystyle=(50-2)(100-1)=5000-50-200+2\equiv2\pmod{125}$

Now as $\displaystyle a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod{m\cdot c}$ where $a,b,m,c$ are integers

$\displaystyle\implies 2^{11}\cdot7^{14}\cdot8\equiv2\cdot8\pmod{125\cdot8}$

Can you take it home from here?

@AlexyVincenzo, please find my other method here – lab bhattacharjee Dec 31 '13 at 18:15 — lab bhattacharjee, Dec 31 '13 at 18:15

Determination of the last three digits of $2014^{2014}$

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