Answers to selected questions in Math Kook (2022)

<hr> Question 1. Start number n=1 is odd, so its successor is (3n+1)/2 = 4/2 = 2.

<hr> Question 2. A number n has two predecessors if n divided by 3 leaves a remainder of 2.

<hr> Question 9. Odd start numbers for the 4n+1 problem remain odd and diverge to infinity.

<hr> Question 17. Better check that again.

<hr> Question 18. All are true.

<hr> Question 22. (143,77) (66,77) (66,11) (11,11)

<hr> Question 24. 3,254,867 / 9,749,241

<hr> Question 30. 2

<hr> Question 31. [14 0]

<hr> Question 35. Any number of the form 2<sup>b</sup> 5<sup>b</sup> c will have b trailing zeroes. 

<hr> Question 38. If m = 4n + 2 (see errata), then m is of the form 2<sup>1</sup> c, where "2" codes for the first position, and "1" codes for the letter "a".

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