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6669Re: hi everyone.

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  • aditsu
    May 1, 2002
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      --- "omidmazaheri" wrote:
      > Hi Every one.
      > I need to solve this problem.please help me.
      >
      > Prove that for each prime number (that is not 2 or 5) and each
      > natural number N ,there is a power of P that ends with 000...001,
      > where the number of zeroes is N-1

      Wow, I solved this problem at a national maths contest when I was
      about 13 y.o. My solution was something like this (it actually works
      for every odd number that doesn't end in 5):

      step 1) each number P of this kind ends in 1, 3, 7 or 9; then P^4
      obviously ends in 1, let q=P^4

      step 2) if q=r*10^s+1 (s>0) then q^2=2*t*10^s+1
      proof: q=r*10^s+1 => q^2=2*(r^2*5*10^(s-1)+r)*10^s+1
      let u=q^2

      step 3) if u=2*t*10^s+1 then u^5=v*10^(s+1)+1
      proof: let w=2*t*10^s; u=w+1 => u^5=w^5+5*w^4+10*w^3+10*w^2+5*w+1;
      we note that w^2 and 5*w are divisible with 10^(s+1), therefore we
      can write u^5=v*10^(s+1)+1 (v is integer)

      step 4) from 2) and 3) we have: q=r*10^s+1 => q^10=v*10^(s+1)+1;
      through induction we can prove that q^(10^N)=r_N*10^(s+N)+1 (where
      r_N are integers) for every N>=0

      step 5) replacing q=P^4 and s=1 we get P^(4*10^N)=r_N*10^(N+1)+1, so
      P^(4*10^N) ends in N zeroes followed by an 1


      final note: this is probably not the shortest solution, but it's an
      elementary one

      Adi
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