klimenkov
  • 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.
Linear Algebra
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schrodinger
  • schrodinger
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ParthKohli
  • 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.
ParthKohli
  • ParthKohli
We want to prove if that is true. :-)
ParthKohli
  • 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.

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ParthKohli
  • ParthKohli
Can we prove this?
klimenkov
  • 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.
ParthKohli
  • ParthKohli
Yes, we did it!
ParthKohli
  • ParthKohli
We have completed a discovery!
ParthKohli
  • ParthKohli
Can we expand this discovery outside three-digit numbers?
klimenkov
  • 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.
ParthKohli
  • ParthKohli
@Preetha Look what we achieved! Thanks to @klimenkov :-)
klimenkov
  • klimenkov
@ParthKohli, @whpalmer4 How can we expand our problem on non 3-digit numbers?
Preetha
  • 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.
dan815
  • dan815
nice :).. however i was wondering about one of your statements, wat about the assumption that ao-a1+a1=n(b+1)
klimenkov
  • klimenkov
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.
dan815
  • dan815
i did not know its such an old post! wow but still pretty awesome
dan815
  • dan815
you have any new proofs?
klimenkov
  • klimenkov
Yes, some day I will post some of them.
dan815
  • dan815
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 a0-a1+a2 just so happens to be b-1
dan815
  • dan815
and cool:) i look forward to it
dan815
  • dan815
is that also not possible being equal?
klimenkov
  • klimenkov
Yes, because \(b-1\) has degree 1. It has to be lower than 1, only 0 degree will be okay.
dan815
  • dan815
ok just making sure thnx

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