The Problem
Below is a problem dealing with number systems:
My Attempt
Since traowo,ptae,tarumpao appear repeatedly and in consistent positions, I assumed they were conjunctions similar to "and" in "two-hundred and thirty".
Thus 1) would be 5 digits, 2) would be 2 digits, 3) would be 1 digit, 4) would be 3 digits, 5) would be 5 digits, 6) would be 2 digits, 7) would be 1 digit, 8) would be 5 digits, 9) would be 3 digits, and 10) would be 5 digits.
Since there are 2 one-digit numbers(naempr and tarumpao), and the smallest ones in the list of cubes are 1 and 8, they must match. And since they are one-digit and the biggest one is 8, b > 8.
There are two two-digit numbers(eser traowo yuow and naempr traowo yempoka), and these must thus correspond to the next smallest cubes, 27 and 64. Take naempr traowo yempoka. Assume that naempr is 8. Then since b > 8, 8b > 64, and naempr traowo yempoka > 8b, but naempr traowo yempoka has to be either 27 or 64 so it is not possible. Thus naempr is 1 and tarumpao is 8.
Consider naempr traowo yempoka. This equals 1*b + c, for some number c < b. The maximum value of c is b - 1, so the maximum value of this number is b + b - 1 = 2b - 1. For this number to be 27 or 64, 2b - 1 cannot be less than 27, so 2b - 1 ≥ 27 → b ≥ 14.
Now consider naempr ptae eser traowo eser. This would be $b^{2} + cb + c$, and so the maximum value of this is $b^{2} + (b - 1)b + (b - 1)$ → $2b^{2} - 1$. Because there are two three-digit numbers, this must be equal to 125 or 216, so it has to be less than 343. But solving the equation $2b^{2} - 1 \space < \space 343$ gives b < 14, which is a contradiction of what we found earlier.
Could someone please explain this to me? I don't understand where I went wrong. Thank you very much.