Using Fermat's Little Theorem to Show Divisibility

We have $$ 5^{10n+8} = 5^{10n} 5^8 = (5^{10})^n 5^8 \equiv 1 \cdot 5^8 \equiv 4 \bmod 11 $$

• multiply by 25 getting $5^{10(n+1)}-100$
• Take remainders using Fermat, getting $1-1\equiv 0\pmod{11}$

Much easier way!

By FLT $5^{10} \equiv 1 \pmod{11}$ so $5^{10n+8}\equiv 5^8$ and $5^{10n +8} -4 \equiv 5^8 -4\pmod {11}$.

So you just have to show that one case the $5^8 \equiv 4 \pmod {11}$. Then every case will be $5^{10n + 8} - 4\equiv 0 \pmod{11}$

Admittedly that requirse calculations but there are 3 ways, each more clever than the other

1) $5^2 = 25\equiv 3 \pmod {11}$. $5^4\equiv 3^2 \equiv 9\equiv -2 \pmod {11}$. $5^8\equiv (-2)^2 \equiv 4 \pmod {11}$.

2) $5^8*5^2 \equiv 5^{10} \equiv 1\pmod {11}$

$5^8*5^2 \equiv 5^8*3 \equiv 1\pmod{11}$ so as $11$ is prime $3^{-1}$ exist as is.... $1 \equiv 12=3*4\pmod{11}$ so $5^8*3*4 \equiv 4\pmod {11}$ and $5^8\equiv 4\pmod {11}$.

3) I'll admit I didn't come up with this.

If $5^8 -4 \equiv A\pmod{11}$ then

$(5^8-4)*25 \equiv A*25\pmod{11}$

$5^{10} - 100 \equiv 3A$

$1 - 1 \equiv 3A$

$3A \equiv 0\pmod {11}$ and as $11$ is primes $A\equiv 0 \pmod{11}$.