How do I convert from binary base to decimal? I have a homework problem and I don't understand it. Here is the problem:
The base two number 11111(base 2) has the same digit in all places. The same 
number can be written in different bases. Find two other bases where the 
number has the same digit in all it's places.
I get that 11111 base 2 = 31, but I don't really understand what the question is asking me or how to "solve" it.
 A: $$11111_2=1\cdot 2^0+1\cdot 2^1+1\cdot 2^2+1\cdot 2^3+1\cdot 2^4=1\cdot 1+1\cdot 2+1\cdot 4+1\cdot 8+1\cdot 16$$
$$=1+2+4+8+16=31$$
A: A number of the form $aaa$ in base $b$ represents $ab^2+ab+a$, with the number of terms matching the number of digits.  You are being asked to find two pairs $a,b$ (with some number of digits) so that it can represent $31_{10}$
A: The number is $31$. If $dd\dots dd_b = 31_{10}$, then $d$ divides $31$.  Only possibilities are $d= 1$ and $d=31$.  So for all bases $b \gt 31$ you have a single digit which clearly passes the requirement.  Now check all $11\dots 11_b$ for different bases $b$ and you have the rest. 
A: The number in decimal is 31.  I think we've established that.
It appears that you're being asked for two other bases in which 31 is written with all the same digits.
Any base 32 or larger will work.  In this case, decimal 31 will be a single digit.  I guess they're all the same if there's only one, right?
Base 30 will work:
$$11_{30}=1\cdot 30^1+1\cdot 30^0$$
Since 31 is odd, the next possibility would be $33$, but that won't work because 28 isn't divisible by 3.
$55$ and higher won't work because the number would need to be at least base 6, and $55_6$ exceeds 31.
Let's go to $111$ then.  Is there a value for $n$ such that $n^2 + n = 30$ ?  Why, yes there is:
$$111_{5}=1\cdot 5^2+1\cdot 5^1+1\cdot 5^0$$
So there are the two (non-trivial single-digit) solutions.
