Animated Solution for Mathematics - Binomial Theorem: The remainder when $(11{)}^{1011}+(1011{)}^{11}$ is divided by 9 is

Have you ever looked at a problem like finding the remainder of $(11{)}^{1011}+(1011{)}^{11}$ divided by $9$ and felt a wave of intimidation? I get it. Those exponents are not just large; they are gargantuan.

But here is the secret that separates the masters from the novices: in mathematics, we do not need to conquer the giant; we only need to understand its cycle. Welcome to the world of modular arithmetic, where we stop caring about the magnitude of a number and start caring about its position on a clock.

Let us isolate the first term. We are looking for $(11{)}^{1011}(\mathrm{mod}9)$. The base is $11$. On our modulo $9$ clock, $11$ is just $2$ steps past $9$, so $11\equiv 2(\mathrm{mod}9)$.

This transforms our expression into $2^{1011}(\mathrm{mod}9)$. We are looking for a power of $2$ that lands us near a multiple of $9$. We know $2^1=2$, $2^2=4$, and $2^3=8$.

There it is! Since $8\equiv -1(\mathrm{mod}9)$, this is our golden ticket. We can rewrite $2^{1011}$ as $(2^3{)}^{337}$. Substituting our discovery, we get:

Since $337$ is an odd number, $(-1{)}^{337}$ remains $-1$. Thus, the first term is congruent to $-1(\mathrm{mod}9)$. To make this a standard positive remainder, we add $9$ to get $8$. So, $(11{)}^{1011}\equiv 8(\mathrm{mod}9)$.

Now for the second term: $(1011{)}^{11}$. We use the classic divisibility rule for $9$: the remainder of a number divided by $9$ is the same as the remainder of the sum of its digits divided by $9$.

The sum of the digits of $1011$ is $1+0+1+1=3$. Therefore, $1011\equiv 3(\mathrm{mod}9)$. Our term becomes $3^{11}(\mathrm{mod}9)$.

This is even simpler. We know $3^2=9$, and $9\equiv 0(\mathrm{mod}9)$. Any power of $3$ greater than or equal to $2$ will contain $3^2$ as a factor, meaning it will be a multiple of $9$. Thus:

The second giant has fallen without a fight!

We have our two pieces: the first term leaves a remainder of $8$, and the second term leaves a remainder of $0$. By the additive property of modular arithmetic, the remainder of the sum is the sum of the remainders:

And there we have it. The remainder is 8. You see? We didn't need a supercomputer. We just needed to look for the patterns hidden within the numbers. Keep practicing this, and soon, you will see these cycles everywhere.