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

    • one year ago
  3. ParthKohli
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    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
  4. ParthKohli
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    Can we prove this?

    • one year ago
  5. klimenkov
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    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
  6. ParthKohli
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    Yes, we did it!

    • one year ago
  7. ParthKohli
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    We have completed a discovery!

    • one year ago
  8. ParthKohli
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    Can we expand this discovery outside three-digit numbers?

    • one year ago
  9. klimenkov
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    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.

    • one year ago
  10. ParthKohli
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    @Preetha Look what we achieved! Thanks to @klimenkov :-)

    • one year ago
  11. klimenkov
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    @ParthKohli, @whpalmer4 How can we expand our problem on non 3-digit numbers?

    • one year ago
  12. Preetha
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    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
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