Basic Bases

The problem is a very interesting one, really cool idea. Anyway, the solution is pretty fun. First, you can write 2707 as a product of powers of 10: $(2\cdot10^{3})+(7\cdot10^{2})+(0\cdot10^{1})+(7\cdot10^{0})$ Replace 10 with our unknown base: $(2\cdot{x^{3}})+(7\cdot{x^{2}})+(0\cdot{x^{1}})+(7\cdot{x^{0}})$ Now, we can set this equal to 9991: $(2\cdot{x^{3}})+(7\cdot{x^{2}})+(0\cdot{x^{1}})+(7)=9991$ And we can turn this into a function: $(2\cdot{x^{3}})+(7\cdot{x^{2}})+(0\cdot{x^{1}})-9894 = 0$ . I just graphed it for simplicity, but y = 0 at x = 16. There fore, $\boxed{16}$ is our answer. Again cool question, sorry about the misunderstanding on my part. :)

9984 = 2700 in our base.

256 * 39 = 2700 in base 16 (39 in base 10 = 27 in base 16)

So the answer is 16