# 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 $b$

Note by Zakir Husain
3 days, 14 hours ago

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.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. 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 1paragraph 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

I think your argument can be adapted to any base, b. The sum would be (b(b-1)/2)^n. Proof could be by induction. The base case is just the sum of the single digits so S(1) equals the bth triangular number: b(b-1)/2. Assuming S(n)=(b(b-1)/2)^n, then by your leading digits argument, S(n+1)=1(S(n))+2(S(n))+...+b(S(n))=(b(b-1)/2)(b(b-1)/2)^n=(b(b-1)/2)^(n+1) completing the proof by induction.

- 15 hours ago

Log in to reply

Good efforts!

- 12 hours ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...