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klimenkov
Group Title
@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.
 one year ago
 one year ago
klimenkov Group Title
@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.
 one year ago
 one year ago

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ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
We want to prove if that is true. :)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Can we prove this?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.7
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.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Yes, we did it!
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
We have completed a discovery!
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Can we expand this discovery outside threedigit numbers?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.7
Proof: 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.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
@Preetha Look what we achieved! Thanks to @klimenkov :)
 one year ago

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

Preetha Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

klimenkov Group TitleBest ResponseYou've already chosen the best response.7
There 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.
 3 months ago

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

dan815 Group TitleBest ResponseYou've already chosen the best response.1
you have any new proofs?
 3 months ago

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

dan815 Group TitleBest ResponseYou've already chosen the best response.1
everything 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
 3 months ago

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

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

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

dan815 Group TitleBest ResponseYou've already chosen the best response.1
ok just making sure thnx
 3 months ago
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