# 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
1 year, 2 months ago

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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.

- 1 year, 2 months ago

Good efforts!

- 1 year, 2 months 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.

- 1 year, 1 month ago