# digit.py
# Let's empirically determine the answers to 3 questions.  
# If you write out the integers from 1 to 1,000,000,
# 1.  How many numbers contain exactly one 5
# 2.  How many numbers contain at least one 5
# 3.  How many 5s would be written
# It doesn't matter if it's "5", but if I said 1 or 0 we'd be a bit sensitive
# about starting at 1 and ending at 1,000,000 instead of 0 to 999,999.
# Since we don't care about appearances of 0, we may assume leading zeros.
# Hopefully, these empirical results should inspire
# the correct procedure for counting these occurrences.
# For 1 to 1 million, the program takes about 1 second to run.

exactCount = 0
atLeastCount = 0
totalCount = 0
max = 1000000

n = 1
while n <= max:
    s = str(n)
    howOften = s.count('5')
    if howOften == 1:
        exactCount += 1
    if howOften >= 1:
        atLeastCount += 1
    totalCount += howOften
    n += 1

print("Numbers from 1 to " + str(max))
print("Containing exactly one 5:   " + str(exactCount))
print("Containing at least one 5:  " + str(atLeastCount))
print("Total 5's written:          " + str(totalCount))


''' various results:
Numbers from 1 to 1000000
Containing exactly one 5:   354294     = 6 * 9^5
Containing at least one 5:  468559     = 10^6 - 9^6
Total 5's written:          600000     = 6 * 10^5

To think about "at least 5", can count indirectly:  Consider if we didn't
want any 5's at all, and subtract from total.  Hence, 10^6 - 9^6.
With more toil, 468559 can also be obtained via inclusion/exclusion:
      C(6,1) * 100,000 =   600,000
    - C(6,2) * 10,000    - 150,000
    + C(6,3) * 1,000     +  20,000
    - C(6,4) * 100       -   1,500
    + C(6,5) * 10        +      60
    - C(6,6) * 1         -       1

Numbers from 1 to 1000
Containing exactly one 5:   243     = 3 * 9^2
Containing at least one 5:  271     = 10^3 - 9^3
Total 5's written:          300     = 3 * 10^2

How do we count 243 containing exactly one 5?
If 5 is the first digit, we only have 9*9 choices for other two digits:  81
And we multiply by the number of digit places the 5 could be in.
In general, our answer is (# digits) * 9 ^ (# digits - 1)
Therefore, for 6 digits, we have 6 * 9^5 = 254294.
'''