For the following multiplicative LCGs, compute Zi for enough values of i ≥ 1 to cover an entire cycle:
(a) Zi = (11Zi – 1)(mod 16), Z0 = 1
(b) Zi = (11Zi – 1)(mod 16), Z0 = 2
(c) Zi = (2Zi – 1)(mod 13), Z0 = 1
(d) Zi = (3Zi – 1)(mod 13), Z0 = 1
Note that (a) and (b) have m of the form 2b; (c) is a PMMLCG, for which a = 2 is a primitive element modulo m = 13.
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