MAT 243 Test 2 SOLUTIONS, FORM A

Transcription

1 MAT 243 Test 2 SOLUTIONS, FORM A 1. [15 points] Calculate the following quantities: a. 17 mod 4 Solution: = = 1. 4 b. 17 div 4 17 Solution: = 4. 4 c. (( 1) mod 12) mod (27 div 5) Solution: (( 1) mod 12) mod (27 div 5) = 11 mod 5 = [10 points] Find real numbers C and k such that 3x 3 + 4x C x 3 whenever x k. Solution: The easiest way to solve this problem is to notice that if x 1, then 3x 3 + 4x x x x 3 + 4x 3 + 8x 3 = 15x 3, so you can let k = 1 and C = 15. Other values were also possible, and can be found by maximizing 3x3 + 4x over the interval [k, ), for some k > 0. (Note that you cannot have k 0, since substituting x = 0 into the desired inequality yields 8 0.) This function is decreasing, so any solution must have C 3+ 4 k + 8 k 3. Grading: done on a point basis. Grading for common mistakes: +3 points (total) for k = 0; +5 points for only giving C or k; +7 points for choosing C 15. x 3 2

2 3. [10 points] Convert 697 to base 7. MAT 243 Test 2 SOLUTIONS, FORM A Solution: Repeatedly divide by 7: 697 = = = = and take the remainders in reverse order: (2014) 7. Grading: Done on a point basis. Grading for common mistakes: +7 points (total) for no work. 4. [15 points] Use mathematical induction to prove that n F 2i = F 2n+1 1 for all n 0, where F n is the Fibonacci sequence (F 0 = 0, F 1 = 1, and F n = F n 1 + F n 2 for all n 2). Solution: Let P (n) be the equation n F 2i = F 2(n)+1 1. P (0) is true, because 0 F 2i = 0 = F 1 1 = F 2(0)+1 1. or Now assume that P (k) is true; that is, F 2i + 1 = F 2(k+1)+1 1. Since F 2i = F 2(k)+1 1. The goal is to prove P (k + 1), k+1 F 2i = F 2i + F 2(k+1) = F 2k F 2k+2 = F 2k+3 1 = F 2(k+1)+1 1, this holds. Thus, by the principle of mathematical induction, P (n) is true for all n 0. Grading: +5 points for the base case, +5 points for the inductive step, +5 points for the conclusion. Grading for common mistakes: +12 points (total) for minor mistakes, +10 points (total) for not being able to complete the inductive step. 3

3 MAT 243 Test 2 SOLUTIONS, FORM A 5. [10 points] Use the Euclidean Algorithm to calculate gcd(843660, 46698). Solution: Okay, I guess you did need a calculator on this test = = = , and the gcd is the last nonzero remainder, or 258. Grading: +9 points for the divisions, +1 points for the answer. Grading for common mistakes: +5 points (total) for finding a prime factorization. 6. [15 points] Let T be the set defined recursively by: (1) 5 T (2) If x T, then 3x T, x + 4 T, and x 8 T (3) The only elements of T are those which are obtained from (1) by applying (2) finitely many times. Prove that for every x T, x is an integer which is not divisible by 4. (Hint: n is not divisible by 4 if n 1 (mod 4), n 2 (mod 4), or n 3 (mod 4).) Solution: First of all, 5 is not divisible by 4; this takes care of the base case. Now suppose that x T and x is not divisible by 4; the goal is to show that 3x, x + 4, and x 8 are not divisible by 4, either. The easiest way to prove this is by considering what x mod 4 is. If x mod 4 = 1, then x 1 (mod 4), and so 3x 3 0, so 3x is not divisible by 4. Also, x + 4 1, so x + 4 is not divisible by 4, either. Lastly, x 8 1, so x 8 is not divisible by 4. Now, if x mod 4 = 2, then x 2 (mod 4). The numbers 3x, x + 4, and x 8 are now (all) equivalent to 2, and so are not divisible by 4. Lastly, if x mod 4 = 3, then x 3 (mod 4), and 3x 1, x + 4 3, and x 8 3. None of these numbers is equivalent to 0 (mod 4), so they are not divisible by 4. This is the last of the cases to consider. Hence, by the principle of structural induction, T does not contain any integers that are divisible by 4. QED. Grading: +5 points for the base case, +5 points for the inductive step, +5 points for the conclusion. This problem was graded very liberally. 4

4 .. MAT 243 Test 2 SOLUTIONS, FORM A 7. [10 points] Give a recursive definition for the set of all (x, y) Z + Z + such that x + y 1 (mod 3). ( Z + is the set of positive integers.) Solution: First of all, 0 is not a positive integer. (That caused quite a few mistakes.) Now, let s draw a picture of some of the elements of this set (which I will call S here.).. From the point (x, y), you can go to the points (x + 3, y),.. (x + 2, y + 1), (x + 1, y + 2), and (x, y + 3) (shown by arrows. above) and still remain in the set S. (Note that if you travel. to (x 1, y + 1), for instance, then you will eventually make the x-coordinate of your point negative; starting at (2, 2), you. would end up at (1, 3), and then (0, 4), which is not in the set.) This gives us the recursive part of the definiton of S. Now you need to determine which points cannot be reached by applying any of the rules above; they turn out to be (1, 3), (2, 2). and (3, 1). Thus, the recursive definition becomes: 1. (1, 3), (2, 2), (3, 1) S. 2. If (x, y) S, then (x + 3, y), (x + 2, y + 1), (x + 1, y + 2), (x, y + 3) S. 3. Only ordered pairs obtained from (1) and using (2) a finite number of times are in S. (Note that the extra ordered pairs (x + 2, y + 1) and (x + 1, y + 2) can be removed as well. There is also symmetry present, so other answers are possible.) Grading: +3 points for the base case, +4 points for the recursive part, +3 points for the finiteness statement. Grading for common mistakes: +5 points (total) for only drawing a picture; 2 points for including points with a zero coordinate; +2 points (total) for not supplying a recursive definition. 8. [15 points] Find a simple function g(n) such that the running time of the following segment of pseudocode is O(g(n)). s 1 k 1 for i from 1 to 3 n do k k + 2 s i k k 1 while k < n do s s + k k 3k s s k Solution: The four assignment statements have a running time of O(1). The for loop has a running time of O( 3 n 1) = O( 3 n), because the loop repeats 3 n times, and every iteration requires time O(1). The while loop repeats at most log 3 n times, so the running time of the while loop is O(log 3 n 1) = O(log 3 n). Lastly, the running time of the entire piece of code is O(max(1, 3 n, log 3 n)) = O( 3 n). Grading for common mistakes: +7 points (total) for O(log n); +10 points for O( 3 n log n); +10 points for O(n) (with n iterations of the while loop); 3 points for not showing any work. 5

5 MAT 243 Test 2 SOLUTIONS, FORM B 1. [15 points] Calculate the following quantities: a. 15 mod 6 Solution: 15 mod 6 = = = 3. 6 b. 15 div 6 Solution: 15 div 6 = 15 = 2. 6 c. (( 2) mod 13) mod (27 div 4) Solution: (( 2) mod 13) mod (27 div 4) = 11 mod 6 = [10 points] Find real numbers C and k such that 7x 4 + 5x C x 4 whenever x k. Solution: The easiest way to solve this problem is to notice that if x 1, then 7x 4 + 5x x x x 3 + 5x 3 + 9x 3 = 21x 3, so you can let k = 1 and C = 21. Other values were also possible, and can be found by maximizing 7x4 + 5x over the interval [k, ), for some k > 0. (Note that you cannot have k 0, since substituting x = 0 into the desired inequality yields 9 0.) This function is decreasing, so any solution must have C 7+ 5 k + 9 k 4. Grading: done on a point basis. Grading for common mistakes: +3 points (total) for k = 0; +5 points for only giving C or k; +7 points for choosing C 21. x 4 2

6 3. [10 points] Convert 442 to base 6. MAT 243 Test 2 SOLUTIONS, FORM B Solution: Repeatedly divide by 6: 442 = = = = and take the remainders in reverse order: (2014) 6. Grading: Done on a point basis. Grading for common mistakes: +7 points (total) for no work. 4. [15 points] Use mathematical induction to prove that n F 2i+1 = F 2n+2 for all n 0, where F n is the Fibonacci sequence (F 0 = 0, F 1 = 1, and F n = F n 1 + F n 2 for all n 2). Solution: Let P (n) be the equation n F 2i+1 = F 2(n)+2. P (0) is true, because 0 F 2i+1 = 1 = F 2(0)+2. Now assume that P (k) is true; that is, F 2i = F 2(k+1)+2. Since F 2i+1 = F 2(k)+2. The goal is to prove P (k + 1), or k+1 F 2i+1 = F 2i+1 + F 2(k+1)+1 = F 2k+2 + F 2k+3 = F 2k+4, this holds. Thus, by the principle of mathematical induction, P (n) is true for all n 0. Grading: +5 points for the base case, +5 points for the inductive step, +5 points for the conclusion. Grading for common mistakes: +12 points (total) for minor mistakes, +10 points (total) for not being able to complete the inductive step. 3

7 MAT 243 Test 2 SOLUTIONS, FORM B 5. [10 points] Use the Euclidean Algorithm to calculate gcd(879231, 72891). Solution: Okay, I guess you did need a calculator on this test = = = , and the gcd is the last nonzero remainder, or 267. Grading: +9 points for the divisions, +1 points for the answer. Grading for common mistakes: +5 points (total) for finding a prime factorization. 6. [15 points] Let T be the set defined recursively by: (1) 4 T (2) If x T, then 2x T, x + 3 T, and x 6 T (3) The only elements of T are those which are obtained from (1) by applying (2) finitely many times. Prove that for every x T, x is an integer which is not divisible by 3. (Hint: n is not divisible by 3 if n 1 (mod 3) or n 2 (mod 3).) Solution: First of all, 4 is not divisible by 3; this takes care of the base case. Now suppose that x T and x is not divisible by 3; the goal is to show that 2x, x + 3, and x 6 are not divisible by 3, either. The easiest way to prove this is by considering what x mod 3 is. If x mod 3 = 1, then x 1 (mod 3), and so 2x 2 0, so 2x is not divisible by 3. Also, x + 3 1, so x + 3 is not divisible by 3, either. Lastly, x 6 1, so x 6 is not divisible by 3. Now, if x mod 3 = 2, then x 2 (mod 3). The numbers 2x, x + 3, and x 6 are now equivalent to 1, 2, and 2, in that order, and so are not divisible by 3. Hence, by the principle of structural induction, T does not contain any integers that are divisible by 3. QED. Grading: +5 points for the base case, +5 points for the inductive step, +5 points for the conclusion. This problem was graded very liberally. 4

8 MAT 243 Test 2 SOLUTIONS, FORM B 7. [10 points] Give a recursive definition for the set of all (x, y) Z + Z + such that x + y 3 (mod 4). ( Z + is the set of positive integers.) Solution: First of all, 0 is not a positive integer. (That caused quite a few mistakes.) Now, let s draw a picture of some of the elements of this set (which I will call S here.).. From the point (x, y), you can go to the points (x + 4, y),.. (x + 3, y + 1), (x + 2, y + 2), (x + 1, y + 3), and (x, y + 4) (shown.. by arrows above) and still remain in the set S. (Note that if you.. travel to (x 1, y+1), for instance, then you will eventually make the x-coordinate of your point negative; starting at (2, 1), you.. would end up at (1, 2), and then (0, 3), which is not in the set.). This gives us the recursive part of the definiton of S. Now you need to determine which points cannot be reached by applying any of the rules above; they turn out to be (2, 1) and. (1, 2). Thus, the recursive definition becomes: 1. (1, 2), (2, 1) S. 2. If (x, y) S, then (x + 4, y), (x + 3, y + 1), (x + 2, y + 2), (x + 1, y + 3), (x, y + 4) S. 3. Only ordered pairs obtained from (1) and using (2) a finite number of times are in S. (Note that the extra ordered pairs (x + 3, y + 1) and (x + 1, y + 3) can be removed as well. There is also symmetry present, so other answers are possible.) Grading: +3 points for the base case, +4 points for the recursive part, +3 points for the finiteness statement. Grading for common mistakes: +5 points (total) for only drawing a picture; 2 points for including points with a zero coordinate; +2 points (total) for not supplying a recursive definition. 8. [15 points] Find a simple function g(n) such that the running time of the following segment of pseudocode is O(g(n)). s 0 k 1 while k n do s s + k 2 k 2k for i from 1 to n DO s i s + 3 s s/k Solution: The three assignment statements have a running time of O(1). The while loop repeats at most log 2 n times, so the running time of the while loop is O(log 2 n 1) = O(log 2 n). The for loop has a running time of O( n 1) = O( n), because the loop repeats n times, and every iteration requires time O(1). Lastly, the running time of the entire piece of code is O(max(1, n, log 2 n)) = O( n). Grading for common mistakes: +7 points (total) for O(log n); +10 points for O( 3 n log n); +10 points for O(n) (with n iterations of the while loop); 3 points for not showing any work. 5