Answers to Problems
Page 266: The first question just needs a count. Remember that the sequence is a cycle, so the last 0 counts as being followed by 0001... taken from the beginning.
The second question can also be done just by trying them out. But it is more useful to see a pattern in the results. We have already found that modulo 17, the powers of 10 are:
10^{1}=10, 10^{2}=15, 10^{3}=14, 10^{4}=4, 10^{5}=6,
10^{6}=9, 10^{7}=5,
10^{8}=16,
10^{9}=7, 10^{10}=2,
10^{11}=3, 10^{12}=13,
10^{13}=11, 10^{14}=8,
10^{15}=12,
10^{16}=1.
So looking at the powers of 15, modulo 17, is the same as looking at the powers of 10^{2}, and clearly these will only give the even powers of 10: 10^{4}, 10^{6}, 10^{8}... 10^{16}.
Similarly for powers of 4, 9, 16, 2, 13, 8. But looking at the powers of 14, modulo 17, is the same as looking at the powers of 10^{3}, and these will run through all 16 possibilities. Similarly for all the odd powers of 10.
The last question is just another way of expressing this observation. The numbers 4, 9, 16, 2, 13, 8 are all squares, modulo 17, and so lie on the diagonal of the base17 multiplication table.
