# Yet yet another google puzzle (last one this week, I swear)

Ok, last one:

```Consider a function which, for a given whole number n,
returns the number of ones required when writing out all numbers
between 0 and n. For example, f(13) = 6. Notice that f(1) = 1.
What is the next largest n such that f(n) = n? ```
Again I failed to solve it with no help from my old good Pentium :( I came up with some sort of formula, which I believe is right, but the number seemed to be quite big, so... 5-lines of code solved it. Can anybody show how it can be deducted in mind?

PS. Oh, forgot to mention - the puzzles were taken from the GLAT.

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This might be easier for some to do this mentally to follow:

f(9) = 1
f(20) = 2*f(9) + 10 = 12 [(1 thru 9) x 2 + 10 (10 thru 19 on the 2nd digit)]

f(99) = 10*f(9) + 10 (counting first digit in 10 thru 19) = 20
f(200) = 2*f(99) + 100 = 140
f(999) = 10*f(99) + 100 = 300
f(2000) = 2*f(999) + 1000 = 1600
f(9999) = 10*f(999) + 1000 = 4000
f(20000) = 2*f(9999)+ 10000 = 18000
f(99999) = 10*f(9999) + 100000 = 50000
f(200000) = 2*f(99999) + 100000 = 200000

And more results. With the last result you have:

f(200000) = 200000

then f(199999) = 200000

If you down the value you have:

if 199991 f(n) = n + 1

the next lower number is:

f(199990) = 199990 (the number of ones is reduced in 2)

if 199981 f(n) = n (great)

and the next lower number is:

f(199980) = 199979 (the number of ones is reduced in 2)

AND THE SOLUTION IS..... 199981

Check the next solution:
f(200) -> 120 + 20
f(2000) -> 1300 + 300
f(20000) -> 14000 + 4000
f(200000) -> 150000 + 50000 (OK Solution)