Finding the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$ 
Find the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by $7$.

Please brief about the concept behind this to solve such problems. Thanks.
 A: The idea is to reach at $\equiv\pm1\pmod n$ for a given modulo integer $n$
In general we should utilize Fermat's little theorem for prime modulo
or Euler's Totient Theorem or Carmichael Function for non-prime modulo 
unless we can reach $\pm1$ easily like below.
$2^3=8\equiv1\pmod 7\implies 2^{100}=2\cdot (2^3)^{33}\equiv2\cdot 1^{33}\pmod 7\equiv2$
$4^3=64\equiv1\pmod 7\implies 4^{100}=4\cdot (4^3)^{33}\equiv4\cdot1^{33}\pmod 7\equiv4$
$3^3=27\equiv-1\pmod 7\implies 3^{100}=3\cdot (3^3)^{33}\equiv3\cdot(-1)^{33}\pmod 7\equiv-3$
$5^3=125\equiv-1\pmod 7\implies 5^{100}=5\cdot (5^3)^{33}\equiv5\cdot(-1)^{33}\pmod 7\equiv-5$
A: Using Euler-Fermat's theorem.
$\phi(7)=6$
$2^{6} \equiv 1 (\mod 7) \implies2^4.2^{96} \equiv 2(\mod7)$
$3^{6} \equiv 1 (\mod 7) \implies3^4.3^{96} \equiv 4(\mod7)$
$4^{6} \equiv 1 (\mod 7) \implies4^4.4^{96} \equiv 4(\mod7)$
$5^{6} \equiv 1 (\mod 7) \implies5^4.5^{96} \equiv 2(\mod7)$
$2^{100}+3^{100}+4^{100}+5^{100} \equiv 5(\mod 7)$
A: Fermat's little theorem says that $a^6\equiv 1 \pmod 7$ whenever $7$ does not divide $a$.  So $2^{100}+3^{100}+4^{100}+5^{100}\equiv 2^4+3^4+4^4+5^4 \pmod 7$
A: The principle is that if $a\equiv a'\pmod n$ and $b\equiv b'\pmod n$, then $ab\equiv a'b'\pmod n$; and similarly for addition instead of multiplication.
A: Try modular arithmetic (http://en.wikipedia.org/wiki/Modular_arithmetic). The remainder you're looking for is the $x$ such that:
$$2^{100}+3^{100}+4^{100}+5^{100}\equiv x\mod7$$
Hint:
If $\cases{2^{100}\equiv a\mod7 \\ 3^{100}\equiv b\mod7 \\ 4^{100}\equiv c\mod7 \\ 5^{100}\equiv d\mod7}$ Then $2^{100}+3^{100}+4^{100}+5^{100}\equiv a+b+c+d\mod7$
A: The answer is 5. Now it may be helpful to work backwards to understand why.
A: 2^100 + 3^100 + 4^100 + 5^100/7 
= (2^3)^33 *2 + (3^3)^33 *3 + (4^3)^33 *3 + (5^3)^33 *3 / 7 
= 2*8^33 + 3*27^33 + 4*64^33 + 5*125^33 / 7
= 2*1^33 + 3*-1^33 + 4*1^33 + 5*-1^33 / 7
= 2*1 + 3*-1 + 4*1 + 5*-1 / 7
= 2 - 3 + 4 -5 / 7 
= -2/7
= 7-2
= 5 remainder
