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Transcript
Math 347 C1 HOUR EXAM II 30 July 2013 SOLUTIONS 1. a) Show that 2561 ≡ 2(mod 17). (Hint: Use the fact that 561 = 3 · 11 · 17 and apply Fermat’s Little Theorem and the Division Algorithm.) b) Suppose that you know additionally that 2561 ≡ 2(mod 3) 2561 ≡ 2(mod 11) 2561 ≡ 2(mod 17) Prove that 2561 ≡ 2(mod 561). SOLUTION a) 2561 = 23·11·17 = (217 )3·11 ≡ 23·11 ≡ 233 (mod 17), since 217 ≡ 2(mod 17) by Fermat’s Little Theorem. By the Division Algorithm, 33 = 17 · 1 + 16 and so 233 = 217+16 = 217 · 216 ≡ 2 · 216 ≡ 217 ≡ 2(mod 17). Hence 2561 ≡ 2(mod 561). b) Consider the system of congruences: x ≡ 2(mod 3) x ≡ 2(mod 11) x ≡ 2(mod 17) Clearly x = 2 is a solution to the system. Moreover, we are given above that 2561 is also a solution to the system. Since the moduli 3, 11 and 17 are pairwise relatively prime, we can apply the Chinese Remainder Theorem which says that any two solutions to the system are congruent modulo the product of the moduli. Thus 2561 ≡ 2(mod 561). OR It follows from the 3 congruences above that 2561 − 2 is divisible by the primes 3, 11 and 17. Hence, by the Fundamental Theorem of Arithmetic, it must be divisible by their product. Therefore 2561 − 2 ≡ 0(mod 561), i..e., 2561 ≡ 2(mod 561). 2. Find all rational solutions to the equation: 2x3 + x2 + x + 2 = 0. Are there any other real solutions to that equation? Prove your answer. SOLUTION According to the Rational Zeros Theorem, if the rational number ab is a solution to the equation above, then a|2 and b|2. Hence the possible solutions are 1, 2, 12 , -1, -2 and − 12 . Since all coefficients of the polynomial are positive, we can ignore the positive numbers in this list. We must therefore test -1, -2, and − 21 . A simple calculation shows that only x = −1 is a solution. Therefore x + 1 is a factor of the polynomial and it is easy to see that 2x3 + x2 + x + 2 = (x + 1)(2x2 − x + 2). To see if 2x3 + x2 + x + 2 = 0 has other real roots, we need to see if 2x2 − x + 2 = 0 has real roots. Using the quadratic formula, we get x = (1 ± √ 1 − 16)/4, which are imaginary numbers. Thus the only real solution to 2x3 + x2 + x + 2 = 0 is x = −1. 3. Consider the congruence equation 49x ≡ 17(mod 149). a) Show that the congruence has a solution. (Hint: Don’t try to calculate a specific solution.) b) Replace one of 49, 17, 149 with another natural number so that the resulting congruence doesn’t have a solution. Explain. SOLUTION a) The congruence has a solution if 49 and 149 are relatively prime. The only prime divisor of 49 is 7 and clearly 7 doesn’t divide 149. Thus 49 and 149 are relatively prime and so the congruence has a solution. b) For example, if we change 149 to 140, i.e., 49x ≡ 17(mod 140), then gcd(49, 140) = 7. Now convert the congruence to the equation: 49x = 17 + 140k. It follows that if a solution exists then since 7 is a divisor of 49 and 140, 7 must divide 17, which is false. Thus 49x ≡ 17(mod 140) has no solution. 4. Let S = [0, 1). a. Find a sequence in S which converges to sup(S). b. Find a sequence in S which converges to inf (S). SOLUTION a) Clearly sup(S) = 1. Let the sequence hai be defined by an = 1 − n1 , for n = 1, 2, . . . . Since 0 ≤ 1 − n1 < 1, each an ∈ S and an → 1 = sup(S). b) Since inf (S) = 0 and 0 ∈ S, we can choose the sequence hbi with bn = 0, for all n and bn → 0 = inf (S).