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 2 years ago
@whpalmer4 asked an interesting question.
If there is only one 3digit number in the base \(b\) the sum of the numbers of which is \(b+1\) times less then the number.
 2 years ago
@whpalmer4 asked an interesting question. If there is only one 3digit number in the base \(b\) the sum of the numbers of which is \(b+1\) times less then the number.

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ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1here'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.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1We want to prove if that is true. :)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1For 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.

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.7This 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.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1We have completed a discovery!

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1Can we expand this discovery outside threedigit numbers?

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.7Proof: Any 3digit number in base \(b\) can be given as \(a_2b^2+a_1b+a_0\). Now divide this number by \(b+1\): \(a_2b^2+a_1b+a_0=(a_2b+(a_1+a_2))(b+1)+(a_0a_1+a_2)\) If we want this number to be divisible by \(b+1\) we need \(a_0a_1+a_2=0\). 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: \(a_2(b^2b1)a_1a_0b=0\) As we know from the previous \(a_1=a_0+a_2\) Put this into the last expression: \(a_2(b^2b2)a_0(b+1)=0\) Can divide by \(b+1\): \(a_2(b2)a_0=0\) That means that \(a_0\) must be divisible by \(b2\). But \(a_0\) can take only values \(0,1,\ldots,b1\). Among them only \(b2\) is divisible by \(b2\). So \(a_0=b2\) Using the last 2 equations \(a_2=1\) And then \(a_1=b1\) The answer: Only one number exists and it is \(n=\overline{1(b1)(b2)}\). That was 198 for b=10. Check this for other b.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1@Preetha Look what we achieved! Thanks to @klimenkov :)

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.7@ParthKohli, @whpalmer4 How can we expand our problem on non 3digit numbers?

Preetha
 2 years ago
Best ResponseYou've already chosen the best response.0This 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.

dan815
 9 months ago
Best ResponseYou've already chosen the best response.1nice :).. however i was wondering about one of your statements, wat about the assumption that aoa1+a1=n(b+1)

klimenkov
 9 months ago
Best ResponseYou've already chosen the best response.7There is a theorem in algebra about dividing polynomials. We can divide polynomials with the reminder. And the degree of the reminder must be lower than the degree of the divider. So it will never happen.

dan815
 9 months ago
Best ResponseYou've already chosen the best response.1i did not know its such an old post! wow but still pretty awesome

dan815
 9 months ago
Best ResponseYou've already chosen the best response.1you have any new proofs?

klimenkov
 9 months ago
Best ResponseYou've already chosen the best response.7Yes, some day I will post some of them.

dan815
 9 months ago
Best ResponseYou've already chosen the best response.1everything makes sense but just 1 more question again if you dont mind xD you said the degree must lower than the divider in this polynomial division but what if a0a1+a2 just so happens to be b1

dan815
 9 months ago
Best ResponseYou've already chosen the best response.1and cool:) i look forward to it

dan815
 9 months ago
Best ResponseYou've already chosen the best response.1is that also not possible being equal?

klimenkov
 9 months ago
Best ResponseYou've already chosen the best response.7Yes, because \(b1\) has degree 1. It has to be lower than 1, only 0 degree will be okay.

dan815
 9 months ago
Best ResponseYou've already chosen the best response.1ok just making sure thnx
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