مسائل نظریه اعداد-مساله 1

  
  صفحه اصلی وبلاگ

دنباله اعداد طبیعی a₁, a₂, a₃, ... را به این صورت تعریف می کنیم که a₁ بر 5 بخش پذیر نیست و a_n+1=a_n+d_n که d_n آخرین رقم a_n است. ثابت کنید این دنباله شامل تعدادنامتناهی جمله به شکل 2^k است.

مساله و حل آنرا به انگلیسی مشاهده فرمایید.

Problem

Define a sequence of of natural numbers a₁, a₂, a₃, ... such that a₁ is not divisible by 5 and a_n+1=a_n+d_n, where d_n is the last digit of a_n. Prove that this sequence contains infinitely many terms of the form 2^k.

Solution

Since d₁ is not 0 and 5, d₂ is equal to 2, 4, 6, or 8. Then the sequence d₂, d₃, ... is periodic with period 4. Therefore, for each n>1

 a_n+4 = a_n +2+4+6+8

and a_n+4p =a_n + 20p for any natural number p. The sequence {a_n} includes a term divisible by 4, say a_n1 =4k. But then a_n1+4p = 4k +20p = 4(k+5p). Since 2^m (mod5) m=0, 1, 2, ... is a periodic sequence 1, 2, 4, 3, 1, 2, 4, 3, ..., infinitely many terms 2^m are equal to k mod(5). Therefore, k+5p contains infinitely many terms of the form 2^m .


http://www.fen.bilkent.edu.tr/~cvmath/Problem/problem-2005.html