wouldn't 503 to base 5 be 2003 there are 20 sets of 5 with no extra sets of 5 and 3 left over correct

I do not understand what you are doing.

If 503 is in base 10
that is 3*10^0 + 0*10^1 + 5*10^2
To put 503 in base 5
first say 503 = 5^x
x log 5 = log 503 (any base logs, use 10)
x = 2.7/.7 = 3.85
so it is 5^3 times something almost 5
1 * 5^3 = 125
2 * 5^3 = 250
3 * 5^3 = 375
4 * 5^3 = 500
we only need 3 more so
3*5^0 + 0*5^1 + 0*5^2 + 4*5^3
or
4 0 0 3 is what I get