In base 10

After writing a solution to this problem I focused on product of digits and got the following result. Also I have a challenge also for all those who found this interesting try if you can do it!

Let $f_{10}(x)$ be defined as the product of digits of $x$ when written in base $10$ for example $f_{10}(279)=2 \times 7 \times 9=126$ . Then, $\boxed{\sum_{i=10^{n-1}}^{10^{n}}f(i)=45^{n}}$

Proof:

First, we have to see the following lemma:

LEMMA : $\sum_{n=10^{q-1}}^{10^{q}}f(n)=45(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))$

Proof:

Let $S_n=\sum_{n=10^{q-1}}^{10^{q}}f(n)$

We can remove numbers from $10^{q-1}$ to $1111...$ ( $q$ $1's$ ) as the numbers between have a $0$ in them

Similarly we have to remove numbers from $2000...$ to $2111...$ ( $q-1$ $0's$ and 1 ), from $3000...$ to $3111...$ ( $q-1$ $0's$ and 1 ) and so on.

$S_n=(1\times1\times1...\times1 )+(1\times1\times1...\times2)...+(9\times9\times9...\times9)$

We can take leading digits common, reducing a single digit from each number

$S_n=1((1\times1...\times1)+(1\times1...2)...+(9\times9...\times9))+2((1\times1...\times1)+(1\times1...2)...+(9\times9...\times9))...9(1\times1...\times1)+9(1\times1...2)...+9(9\times9...\times9))$

Now we can take $\sum_{n=10^{q-2}}^{10^{q-1}}$ by include numbers like $100...01$ as $f(100...01)=0$ so it makes no change

$S_n=(\sum_{n=1}^{9})(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))=45(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))$

Using the lemma we can get

$\sum_{i=10^{n-1}}^{10^{n}}f(i)=45(\sum_{i=10^{n-2}}^{10^{n-3}}f(i))$

If you apply it again and again

$\sum_{i=10^{n-1}}^{10^{n}}f(i)=(45)(45)...(\sum_{i=10^{n-n}}^{10^{1}}f(i))=(45)(45)...(45)=\boxed{45^n}$ Hence proved

Challenge:

I have proved this for base $10$ you can try for any other base or you may prove it for any base $b$

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold


- bulleted
- list

bulleted
list


1. numbered
2. list

numbered
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"

Math

Appears 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}

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered 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"

Math	Appears 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

I think your argument can be adapted to any base, $b$ . The sum would be $\left ( \frac {b(b-1)}{2} \right )^\textup{n}$ .

The proof could be done by induction. The base case is just the sum of the single digits so $\mathbf{S}_{1}$ equals the $(b-1)th$ triangular number: $\frac{b(b-1)}{2}$ .

Assuming $\mathbf{S}_{n}\textup{=}\left ( \frac{b(b-1)}{2} \right )^\textup{n}$ , then by your leading digits argument, $\mathbf{S}_{n+1}=1(\mathbf{S}_{n})+2(\mathbf{S}_{n})+...+(b-1)(\mathbf{S}_{n})=\frac{b(b-1)}{2}(\mathbf{S}_{n})=\left ( \frac{b(b-1)}{2}\right )\left ( \frac{b(b-1)}{2} \right )^\textup{n}=\left ( \frac{b(b-1)}{2} \right )^\textup{n+1}$ , completing the proof by induction.

Justin Travers - 1 year ago

Good efforts!

Zakir Husain - 1 year ago

The problem you mentioned is super, and generalising is even more elegant. I need to try solving the problem you linked before studying this note.

Mahdi Raza - 1 year ago

Math	Appears 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}`	$\boxed{123}$