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 f10(x)f_{10}(x) be defined as the product of digits of xx when written in base 1010 for example f10(279)=2×7×9=126f_{10}(279)=2 \times 7 \times 9=126. Then, i=10n110nf(i)=45n\boxed{\sum_{i=10^{n-1}}^{10^{n}}f(i)=45^{n}}


First, we have to see the following lemma:

LEMMA : n=10q110qf(n)=45(i=10q210q1f(i))\sum_{n=10^{q-1}}^{10^{q}}f(n)=45(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))


Let Sn=n=10q110qf(n)S_n=\sum_{n=10^{q-1}}^{10^{q}}f(n)

We can remove numbers from 10q110^{q-1} to 1111...1111... (qq 1s1's) as the numbers between have a 00 in them

Similarly we have to remove numbers from 2000...2000... to 2111...2111... (q1q-1 0s0's and 1 ), from 3000...3000... to 3111...3111... (q1q-1 0s0's and 1 ) and so on.

Sn=(1×1×1...×1)+(1×1×1...×2)...+(9×9×9...×9)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


Now we can take n=10q210q1\sum_{n=10^{q-2}}^{10^{q-1}} by include numbers like 100...01100...01 as f(100...01)=0f(100...01)=0 so it makes no change


Using the lemma we can get


If you apply it again and again

i=10n110nf(i)=(45)(45)...(i=10nn101f(i))=(45)(45)...(45)=45n\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


  • I have proved this for base 1010 you can try for any other base or you may prove it for any base bb

Note by Zakir Husain
2 months, 3 weeks ago

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1 vote

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I think your argument can be adapted to any base, bb. The sum would be (b(b1)2)n\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 S1\mathbf{S}_{1} equals the (b1)th(b-1)th triangular number: b(b1)2\frac{b(b-1)}{2}.

Assuming Sn=(b(b1)2)n\mathbf{S}_{n}\textup{=}\left ( \frac{b(b-1)}{2} \right )^\textup{n}, then by your leading digits argument, Sn+1=1(Sn)+2(Sn)+...+(b1)(Sn)=b(b1)2(Sn)=(b(b1)2)(b(b1)2)n=(b(b1)2)n+1\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 - 2 months, 2 weeks ago

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Good efforts!

Zakir Husain - 2 months, 2 weeks ago

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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 - 2 months ago

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