Shifting Digits and a Point of View

The number xn is defined as the last digit in the decimal representation of the integer ⌊(2)n⌋ (n = 1, 2, ...). Determine whether the sequence x1, x2, ..., xn, ... is periodic. [Savchev & Andreescu, p. 146].

(⌊a⌋ is the whole part of number a which is more often at this site and elsewhere denoted [x].)

Solution

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

Observe that

 y2n+1= (2)2n+1
  = (2)×2n.

Recollect that in the decimal system the multiplication of a real number by 10 causes the decimal point (or comma in some cultures) to shift one position to the right. The same happens in the binary system when a number is multiplied by 2. It follows that y2n+1 is a digit in the binary representation of 2 so that 0.y1y3y5... is the binary representation of the fractional part of 2. Since 2 is irrational, the sequence y1, y3, y5, ... is not periodic.

Let xn = zn (mod 2), where zn is either 0 or 1, n = 1, 2, ... Clearly, z2n+1 = y2n+1, and we just proved that the sequence {y2n+1} is not periodic. We may now claim that neither is {zn}. For if it were periodic with, say, period N, then it would be periodic with the period 2N as well. But this would imply that the sequence {y2n+1} had a period of N, which is impossible. Therefore, the sequence of remainders {zn} of the division of the terms {xn} by 2 is not periodic. Thus {xn} could not be periodic either.

References

  1. S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

71491799