$\left (14^{2014} \right )^{2014} \mod 60$ without a calculator

Question

Calculate without a calculator:

$\left (14^{2014} \right )^{2014} \mod 60$

I was trying to solve this with Euler's Theorem, but it turned out that the gcd of a and m wasn't 1.

This was my solution so far (also doesn't really work without a calculator):

Prime factorization of 60:
$60 = 2^{2}\cdot 3\cdot 5$

Chinese Remainder Theorem setup:
$\left (14^{2014} \right )^{2014} \mod 4\\$
$\left (14^{2014} \right )^{2014} \mod 3\\$
$\left (14^{2014} \right )^{2014} \mod 5\\$

Euler's Theorem:
$\phi \left ( 4 \right ) = 2$
$\Rightarrow 14^{2}\equiv 1 \mod 4$

Whoops, that's not right because gcd(14,4) is not 1, but 2. Euler's Theorem only works when gcd(a,m) = 1.

I don't know how else to solve this, does anyone have an idea?

Answer 1

$$\left (14^{2014} \right )^{2014}=(14^2)^{1007\times2014}\equiv 16^{1007\times2014} \mod 60$$

Next, notice that $16\times 16\equiv 16\pmod{60}$ to conclude $16^n\equiv 16\pmod{60}$ for all integers $n\ge 1$

Answer 2

$$X=\large (14^{2014})^{2014}=14^{2014^2}$$

Now, $60=3\times 4\times 5$. Now,

$$X\equiv\begin{cases}(-1)^{2014^2}\equiv 1\pmod3\\ (-1)^{2014^2}\equiv 1\pmod5\\ 7^{2014^2}\cdot 4^{(2014^2)/2}\equiv 0\pmod4\end{cases}$$

Now, use Chinese Remainder Theorem.

Answer 3

With all the powers of $2$, $(14^{2014})^{2014}$ is clearly equivalent to $0\pmod4$. We also have $(14^{2014})^{2014}\equiv[(-1)^{2014}]^{2014}\equiv 1\pmod{15}$. These $2$ conditions combined yield a result equivalent to $16\pmod{60}$.

Answer 4

As $(a^m)^n=a^{mn},$

we need $14^{2014^2}\pmod{60}$

As $(14^n,60)=2^2$ for integer $n\ge2,$

let us compute $14^{2014^2-2}\pmod{15}$

Now $14\equiv-1\pmod{15}\implies14^{2014^2-2}\equiv(-1)^{2014^2-2}\pmod{15}\equiv1$ as $2014^2-2$ is even

$\implies14^2\cdot14^{2014^2-2}\equiv14^2\cdot1\pmod{14^2\cdot15}$

As $60|14^2\cdot15,$
$$14^2\cdot14^{2014^2-2}\equiv14^2\pmod{60}\equiv ?$$

Answer 5

Carmichael's function for $60$ is lcm$(\phi(4),\phi(3),\phi(5))=$ lcm$(2,2,4)=4$, so $a^k \equiv a^{k+4}$ once all possible prime factors of $60$ have been accumulated. So we can quickly square $14$ twice and see where it's going: $$ 14^2 = 196 \equiv 16 \bmod 60 \\ 14^4 \equiv 16^2 = 256 \equiv 16 \bmod 60 $$ So we are fortunate here, and $14^{2k} \equiv 16^k \equiv 16 \bmod 60$. No need to worry about the second exponentiation, except to observe that the resulting exponent is certainly even.

Answer 6

By below $\ 2\mid 15-1\,\ \Rightarrow\,\ 14^{2k}\,\equiv\, 4(4^{-1}\,{\rm mod}\ 15)\equiv 4^2\!\pmod{4^2\!\cdot 15},\,\ j = 2\le 2k$.

Lemma $\,\ a\mid n\!-\!1\,\Rightarrow\,(n\!-\!1)^{2k}\equiv a^j(a^{-j}\ {\rm mod}\ n)\ \pmod{a^jn}\, $ if $\ 0\le \color{#c00}{j\le 2k}$

Proof $\ $ Note $\, {\rm mod}\ n\!:\ (n\!-\!1)^{2k}\equiv (-1)^{2k}\equiv 1\equiv a^j(a^{-j}\ {\rm mod}\ n)$

$a\mid n\!-\!1\!\!\underset{\color{#c00}{j\,\le\, 2k}}\Rightarrow\! {\rm mod}\ a^j\!:\ (n\!-\!1)^{2k}\color{#c00}{\equiv 0}\equiv a^j(a^{-j}\ {\rm mod}\ n).\,$ Being congruent both mod $\,a^j\,$ & $\,n,\,$ they are also congruent mod $\,{\rm lcm}(a^j,n) = a^jn,\,$ by $\,(a,n)=1,\,$ by $\,a\mid n\!-\!1.$