Waste less time on Facebook — follow Brilliant.
×

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
5 months, 4 weeks ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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,

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

Prasun Biswas - 5 months, 4 weeks ago

Log in to reply

What do you mean, by saying "fixing" ?

Fidel Simanjuntak - 5 months, 4 weeks ago

Log in to reply

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

Fidel Simanjuntak - 5 months, 4 weeks ago

Log in to reply

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 \(f(n)=k+f(n-1)\).

Prasun Biswas - 5 months, 4 weeks ago

Log in to reply

@Prasun Biswas What do you mean, by saying "fixing" ?

Fidel Simanjuntak - 5 months, 4 weeks ago

Log in to reply

@Fidel Simanjuntak 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.

Prasun Biswas - 5 months, 4 weeks ago

Log in to reply

@Prasun Biswas 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?

Fidel Simanjuntak - 5 months, 4 weeks ago

Log in to reply

@Fidel Simanjuntak 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.

Prasun Biswas - 5 months, 4 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...