Calculating the value with large exponents

Question

I'm trying to solve $2014^{2015}$ $\pmod {11}$, is there a trick or tip to break the problem down to make it easier to solve?

Answer 1

Your result will have $6658$ digits in base $10$. Alpha seems up to the task, but I got tired of clicking on More Digits. Are you sure you don't want this modulo some number?

Now that you added the $\pmod {11}$, it becomes much easier. Note that $2014 \equiv 1 \pmod {11}$

Answer 2

If you are trying to do this efficiently on a computer or something, you can use a binary expansion of $2015$ to calculate the number.

$$2015 = 1 + 2 + 4 + 8 + 16 + 64 + 128 + 256 + 512 + 1024$$

thus

$$x^{2015} = x\cdot x^2\cdot x^4\cdot x^8\cdot x^{16}\cdot x^{64}\cdot x^{128}\cdot x^{256}\cdot x^{512}\cdot x^{1024}$$

This method is much faster than $x \cdot x \cdot x \cdots $.

Answer 3

It's important to remember that if $a \equiv b \mod m$ then $a \cdot c \equiv b \cdot c \mod m$. From this you can prove if $a \equiv b \mod m$ then $a^c \equiv b^c \mod m$.

So what is the remainder of 2014 divided by 11?