Show that an integer in an odd base system is odd in the base 10 system if and only if it has an odd number of odd digits.
I have an idea of how the proof should go in my head, but how do I express the number in the odd base system? Would this be appropriate?
Suppose a number abc in base d, where d is an odd base. Let a,b and c be odd digits. Then in base 10, abc=ad^2+bd+c Let a=2j+1, b=2k+1, c=2m+1 and d=2n+1 Thus, abc=ad^2+bd+c=(2j+1)(4n^2+4n+1)+(2k+1)(2n+1)+(2m+1)=2(4j(n^2)+2(n^2)+4nj+2kn+3n+j+k+m+1)+1 which is odd. Then repeat but use two of the digits as odd and one as even.
Is there a better way to generalize this proof?
You have the basic idea. You can represent an $n$ digit number in base $d$ as $\sum_{i=0}^{n-1}a_id^i$, where $a_i$ are the digits. I would then argue $\pmod 2$, so $d \equiv 1 \pmod 2$ and $\sum_{i=0}^{n-1}a_id^i\equiv \sum_{i=0}^{n-1}a_i \pmod 2$ Now if you know the sum of an odd number of odd numbers is odd you are home.