Animated Solution for Mathematics - Binomial Theorem: The remainder on dividing $5^{99}$ by 11 is ______.

The expression $5^{99}(\mathrm{mod}11)$ appears daunting due to the magnitude of the exponent. However, in modular arithmetic, we focus on the cyclic nature of remainders rather than the absolute value of the number.

We begin by calculating the powers of $5$ modulo $11$ to identify a repeating cycle:

$5^2=25\equiv 3(\mathrm{mod}11)$ (since $25=2\times 11+3$)

$5^3=3\times 5=15\equiv 4(\mathrm{mod}11)$ (since $15=1\times 11+4$)

$5^4=4\times 5=20\equiv 9(\mathrm{mod}11)$ (since $20=1\times 11+9$)

$5^5=9\times 5=45\equiv 1(\mathrm{mod}11)$ (since $45=4\times 11+1$)

We have reached the identity element, $1$. This signifies that the cycle of remainders $\{5,3,4,9,1\}$ repeats every $5$ steps.

By Fermat's Little Theorem, we know that for a prime $p$ and an integer $a$ not divisible by $p$, $a^{p-1}\equiv 1(\mathrm{mod}p)$. Here, $5^{10}\equiv 1(\mathrm{mod}11)$, which confirms our cycle length is a divisor of $10$.

To evaluate $5^{99}(\mathrm{mod}11)$, we determine how many full cycles of $5$ fit into the exponent $99$:

This allows us to rewrite the expression as:

Since $5^5\equiv 1(\mathrm{mod}11)$, the expression simplifies significantly:

The final remainder is $9$. This demonstrates the elegance of modular arithmetic in reducing complex problems to simple, repeating patterns.