## klimenkov @whpalmer4 asked an interesting question. If there is only one 3-digit number in the base $$b$$ the sum of the numbers of which is $$b+1$$ times less then the number. one year ago one year ago

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1. ParthKohli

here's what we noticed: There is only one such number in base $$b$$ such that its sum of the number's digits is $$\dfrac{1}{b + 1}$$ of the number.

2. ParthKohli

We want to prove if that is true. :-)

3. ParthKohli

For example, in base $$10$$, there is only one such number such that the sum of its digits is $$\dfrac{1}{10 + 1}$$ the number: 198. In base 16, he told us that there is only one such number which has its sum of digits equal to $$\dfrac{1}{17}$$ the number.

4. ParthKohli

Can we prove this?

5. klimenkov

This is the beginning of this problem. @ParthKohli solved it. I wrote a code in Mathematica and got this. $$b=2\quad n=6=110_2$$ $$b=3\quad n=16=121_3$$ $$b=4\quad n=30=132_4$$ $$b=5\quad n=48=143_5$$ $$b=6\quad n=70=154_6$$ $$b=7\quad n=96=165_7$$ $$b=8\quad n=126=176_8$$ $$b=9\quad n=160=187_9$$ $$b=10\quad n=198=198_{10}$$ $$b=11\quad n=240=\text{1a9}_{11}$$ $$b=12\quad n=286=\text{1ba}_{12}$$ $$b=13\quad n=336=\text{1cb}_{13}$$ $$b=14\quad n=390=\text{1dc}_{14}$$ $$b=15\quad n=448=\text{1ed}_{15}$$ $$b=16\quad n=510=\text{1fe}_{16}$$ For any base there is only one number.

6. ParthKohli

Yes, we did it!

7. ParthKohli

We have completed a discovery!

8. ParthKohli

Can we expand this discovery outside three-digit numbers?

9. klimenkov

Proof: Any 3-digit number in base $$b$$ can be given as $$a_2b^2+a_1b+a_0$$. Now divide this number by $$b+1$$: $$a_2b^2+a_1b+a_0=(a_2b+(a_1+a_2))(b+1)+(a_0-a_1+a_2)$$ If we want this number to be divisible by $$b+1$$ we need $$a_0-a_1+a_2=0$$. Next one is the statement of the problem: $$(b+1)(a_2+a_1+a_0)=a_2b^2+a_1b+a_0$$ Change this a little: $$a_2(b^2-b-1)-a_1-a_0b=0$$ As we know from the previous $$a_1=a_0+a_2$$ Put this into the last expression: $$a_2(b^2-b-2)-a_0(b+1)=0$$ Can divide by $$b+1$$: $$a_2(b-2)-a_0=0$$ That means that $$a_0$$ must be divisible by $$b-2$$. But $$a_0$$ can take only values $$0,1,\ldots,b-1$$. Among them only $$b-2$$ is divisible by $$b-2$$. So $$a_0=b-2$$ Using the last 2 equations $$a_2=1$$ And then $$a_1=b-1$$ The answer: Only one number exists and it is $$n=\overline{1(b-1)(b-2)}$$. That was 198 for b=10. Check this for other b.

10. ParthKohli

@Preetha Look what we achieved! Thanks to @klimenkov :-)

11. klimenkov

@ParthKohli, @whpalmer4 How can we expand our problem on non 3-digit numbers?

12. Preetha

This is awesome. I can't say I understand it, but it is a great example of collaborative learning, where together you achieved something. Thanks so much for sharing.