The value $2^8 -1$ is divisible by 3 prime numbers. What is the sum of the three prime numbers?
To find the prime factors of $2^8 -1$, I could try to factor it by using the difference of squares formula. That is, I can write $2^8 -1$ as $(2^4)^2 - 1^2$, which is $(2^4 + 1)(2^4 - 1)$. Now, I can see that $2^4 + 1 = 17$, which is a prime number. The other factor, $2^4 - 1$, can be factored further by using the difference of squares formula again. That is, I can write $2^4 - 1$ as $(2^2)^2 - 1^2$, which is $(2^2 + 1)(2^2 - 1)$. Now, I can see that $2^2 + 1 = 5$, which is another prime number. The other factor, $2^2 - 1$, is $3$, which is also a prime number. So, I have found the three prime factors of $2^8 -1$: they are 3, 5, and 17. The sum of the three prime factors is $3 + 5 + 17 = 25$. 