Question 23.85

55. In doubling on a “Stone Age” calculator (see Example 12), both sequences and also the sequence in the Self Check question all end up in the same orbit of attraction, or loop.

  1. What happens if we start with 39?
  2. Can you find a starting value that is not “attracted” to the indicated orbit?
  3. image Explain what you observe in parts (a) and (b) give a general argument about why every starting value must eventually lead to a repeating orbit.

55.

(a) The last entry shown for the first sequence is the fourth entry of the second sequence, so the first “joins” the second and they then both end up going through the same cycle (loop) of numbers over and over.

(b) 39, 78, 56, and we have “joined” the second sequence. However, an initial 00 stays 00 forever; and any other initial number ending in 0 “joins” the loop sequence 20, 40, 80, 60, 20, ….

(c) Regardless of the original number, after the second push of the key, we have a number divisible by 4 and all subsequent numbers are divisible by 4. There are 25 such numbers between 00 and 99. You can verify that an initial number either joins the self-loop 00 (the only such numbers are 00, 50, and 25); joins the loop 20, 40, 80, 60, 20, … (the only such numbers are the multiples of 5 other than 00, 50, and 25); or joins the big loop of the other 20 multiples of 4.