# I really Need a help!

For every real positive integer $$n$$, define $$f(n)$$ as the number of $$1$$'s that appear(s) in the decimal representation of all positive integer numbers from $$1$$ to $$n$$. For example, $$f(6) =1$$ and $$f(16) = 9$$. Prove that $$f(10^n) = n(10^{n-1}) + 1$$ for all positive integer $$n$$.

Note by Fidel Simanjuntak
1 year, 4 months ago

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There is a simple combinatorial argument for this.

Notice that $$f(n)=k+f(n-1)$$ where $$k$$ is the number of 1s in $$n$$. Now, we have only a single 1 in powers of 10, so $$f(10^n)=f(10^n-1)+1$$. Now, all numbers $$\leq 10^n-1$$ are at most $$n$$ digit numbers (can be thought of as $$n$$ digit numbers using non negative digits). Now, we need to count the numbers having 1 in their digits. Fixing 1 at one of the digits places is done in $$n$$ ways and the other $$n-1$$ digit places can be assigned a digit from 0 to 9 in 10 ways for each digit place, hence $$10^{n-1}$$ ways. Notice that there's no problem with excessive counting or missing out on counting a few since if a number has $$m$$ 1s, it is counted exactly once for fixing 1 at each of the $$m$$ digit places, thus it getting counted exactly $$m$$ times as desired. Now then, by the rule of product, we have $$f(10^n-1)=n10^{n-1}$$ and hence we conclude that,

$f(10^n)=n\cdot 10^{n-1}+1$

- 1 year, 4 months ago

What do you mean, by saying "fixing" ?

- 1 year, 4 months ago

Can you give a proof for $$f(n) = k + f(n-1)$$? I still don't get it..

- 1 year, 4 months ago

If $$f(n)$$ is the number of 1s in the numbers from 1 to $$n$$, then this is equivalent to the number of 1s in the numbers from 1 to $$n-1$$ plus the number of 1s in $$n$$, right? Denoting by $$k$$ the number of 1s in $$n$$, we can write $$f(n)=k+f(n-1)$$.

- 1 year, 4 months ago

What do you mean, by saying "fixing" ?

- 1 year, 4 months ago

By "fixing", I mean specifying one of the digit places to be 1 (hence making the digit place fixed) and letting the other digit places vary.

Such "fixing" argument tend to be common in counting (enumeration) problems.

- 1 year, 4 months ago

I'm sorry for bothering you, but I still don't get this " Notice that there's no problem with excessive counting or ........". Can you help me?

- 1 year, 4 months ago

That part was to show that if a number has $$m$$ 1s, it is getting counting exactly $$m$$ times, no more no less. This is because we want to count the number of occurrences of 1s for that number, which is $$m$$.

And don't worry, if you have any further problems understanding my solution, feel free to ask. I'll try my best to help you understand. After all, a solution is no good if it can't be completely understood by the reader. ;)

PS: I'm typing from my mobile, so I might be late responding to your queries, sorry for that.

- 1 year, 4 months ago