Find the remainder when $24^{50} - 15^{50}$ is divided by 13.
This problem involves some modular arithmetic, which is a way of working with remainders after division. For example, if I say that $a \equiv b \pmod{13}$, that means that $a$ and $b$ have the same remainder when divided by 13. One useful property of modular arithmetic is that if $a \equiv b \pmod{13}$ and $c \equiv d \pmod{13}$, then $a + c \equiv b + d \pmod{13}$ and $a \times c \equiv b \times d \pmod{13}$. This means that we can simplify expressions by replacing numbers with their remainders, as long as we are only adding and multiplying them. So, let's look at $24^{50} - 15^{50}$ and see what remainders we can find. First, notice that $24 \equiv 11 \pmod{13}$, since 24 has a remainder of 11 when divided by 13. Similarly, $15 \equiv 2 \pmod{13}$. This means that $24^{50} - 15^{50} \equiv 11^{50} - 2^{50} \pmod{13}$. Now, we need to find the remainders of $11^{50}$ and $2^{50}$ when divided by 13. One way to do this is to notice that $11 \equiv -2 \pmod{13}$, since 11 and -2 have the same remainder. This means that $11^{50} \equiv (-2)^{50} \pmod{13}$. But $(-2)^{50} = 2^{50}$, since the negative sign cancels out when we raise it to an even power. So, we have $11^{50} \equiv 2^{50} \pmod{13}$. This means that $11^{50} - 2^{50} \equiv 0 \pmod{13}$. Therefore, the remainder when $24^{50} - 15^{50}$ is divided by 13 is 0. 