Unformatted text preview: = 412. (b) f(k 2 + 40) = 40 2 + 80k 2 + k4 + 40+ k 2 + 41 = k4 + 81k 2 + 412 = (k 2 + 41)2 k 2 = (k 2 + 41 + k)(k 2 + 41 k). 19. The answer is no, since 333333331 = 19607843 x 17. Exercises 1.1 1. (a) [BB] p q .q (.q) V P p/\((.q)Vp) T T F T T T F T T T F T F F F F F T T F (b) p q .p (.p) + q p/\q (p/\q) V ((&quot;:&quot;p) + q) T T F T T T T F F T F T F T T T F T F F T F F F (c) p q qVp p/\ (qV p) .(p/\(qVp)) . (p/\ (qV p)) +t P T T T T F F T F T T F F F T T F T F F F F F T F (d) [BB] p q r .q pV (.q) . (p V (.q)) .p (.p) V r (. (p V (.q))) /\ ((.p) V r) T T T F T F F T F T F T T T F F T F F T T F F T T T T F F T T T F T T F T T F F T F F F F T F F T T F F F F F T F F F T T T T F F F T T F T T F...
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 Summer '10
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 Graph Theory, Integers, Prime number, 2  k, 80k, standard checker board, 41  k, 81k

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