ELEC-270 Solutions to Assignment 5

Transcription

1 ELEC-270 Solutions to Assignment 5 1. How many positive integers less than 1000 (a) are divisible by 7? (b) are divisible by 7 but not by 11? (c) are divisible by both 7 and 11? (d) are divisible by 7 or 11? (e) are divisible by exactly one of 7 and 11? (f) are divisible by neither 7 nor 11? (a) Every seventh number is divisible by 7. Therefore there are 999 / such numbers. Note that we use the floor function, because the k th multiple of 7 does not occur until the number 7k has been reached. (b) For solving this part and the next four parts, we need to use the Principal of Inclusion-Exclusion. As in (a), there are floor 999 / numbers in our range that are divisible by both 7 and 11 (meaning they are divisible by 77). If we remove these 12 numbers from the 142 numbers divisible by 7, we are left with those that are divisible by 7 but not 11, i.e., there are =130 of them. (c) As explained in (b), there are 12 of them. (d) As in (a), there are floor 999 /11 90 numbers divisible by 11. By the Principle of Inclusion-Exclusion, there are = 220 that are divisible by 7 or 11. (e) We need to subtract from the answer to (d) the number calculated in (c). I.e., there are that are divisible by exactly one of 7 and 11. (f) If we subtract the answer to (d) from the total number of positive integers less than 1000, we have the number divisible by neither 7 nor 11, i.e., = Let A = {w, x, y, z } and B = {1, 2, 3}. (a) How many functions are there from A to B? (b) i. How many functions are there from A to {1, 2}? ii. How many functions are there from A to {2, 3}? iii. How many functions are there from A to {1, 3}? (c) A function f: A B is called onto if for all b B there is at least one a A with f (a) = b. I.e., every element in the range B gets mapped to by at least one element in the domain A. i. Since none of the functions in b(i), (ii), (iii) would be onto functions from A to B (because in each case, some element of B is left out of the range of values), how many functions are there from A to B that are not onto? Be careful not to count a function more than once. E.g., the function that maps w to 2, x to 2, y to 2 and z to 2 is a function in both b(i) and b(ii). ii. What principle are you using in c(i)? (d) Using the results from (a) and (c)(i), how many onto functions are there from A to B?

2 (a) Each of the 4 elements of A may get mapped to one of the 3 elements of B. there are 3 4 possible functions from A to B. (b) Each of the 4 elements of A may get mapped to one of the 2 elements in the subset of B. for each (i), (ii), (iii), get 2 4. (c) (i) Now, none of the 32 4 functions in (b)(i) are onto. Subtract ones that are not distinct. The only functions (i) and (ii) share are those that map each element of A to {2} (because any other function f of (i) maps some elements of A to {1} and such an f would be different from a function of (ii) that can t map anything to {1}) 3 functions must be subtracted from 32 4 Total number of functions that are not onto: (ii) Principle of Inclusion-Exclusion (d) Total # of onto functions: 3 4 (32 4 3) = How many bit strings of length 10 contain five consecutive 0s or five consecutive 1s? First we count the number of bit strings of length 10 that contain five consecutives 0 s. We will base the count on where the string of five or more consecutive 0 s starts. If it starts in the first bit, then the first five bits are all 0 s, but there is free choice for the last five bits; therefore there are 2 5 = 32 such strings. If it starts in the second bit, then the first bit must be a 1, the next five bits are all 0 s, but there is free choice for the last four bits; therefore there are 2 4 = 16 such strings. If it starts in the third bit, then the second bit must be a 1, but the first bit and the last three bits are arbitrary; therefore there are 2 4 = 16 such strings. Similarly, there are 16 such strings that have the consecutive 0 s starting in each of positions four, five, and six. This gives us a total of = 112 strings that contain five consecutive 0 s. Symmetrically, there are 112 strings that contain five consecutive 1 s. Clearly there are exactly two strings that contain both ( and ). Therefore by the inclusion-exclusion principle, the answer is = Show that if five points are picked in the interior of a square with a side of length 2, then at least two of these points are no farther than 2 apart. Divide the interior of the square, with lines joining the midpoints of opposite sides, into four 1 1 squares as shown below. By the pigeonhole principle, at least two of the five points must be in the same small square. The furthest apart two points in a square could be is the length of the diagonal, which is 2 for a square 1 unit on a side

3 Note: the most common mistake people usually make in trying to prove the above result is an (incorrect) argument about how the points are distributed. First of all, you cannot choose how the 5 points are to be distributed. Incorrect reasoning would include something along the lines of the following: If you distribute the points as far apart as possible (say, on the four corners of the square), then obviously this is the worst distribution you can choose and even in this case, it s obvious that any 5 th point thrown into the square will have to be at least 2 distance from some other point (since if the 5 th point is in the middle of the square then it s 2 from each corner point and if it s not in the middle then it s closer than 2 from some corner point). The foregoing reasoning is incorrect because it assumes a particular distribution under the assumption that this distribution is worse than all others. This notion of worse is a sort of feeling about the distribution and is not mathematically precise: there is nothing worse or better about this particular distribution than any other and it, therefore, doesn t prove anything about all possible distributions. 5. (Adapted from Textbook exercise 16, Supplementary Exercises of Chapter 6) (a) Show that in any set of n+1 positive integers not exceeding 2n there must be two that are relatively prime. (b) Show, by example, that if 100 integers are selected from the set {1, 2, 3,, 200} then it is possible that no two numbers in the selection are relatively prime. (a) Partition the set of numbers 1 to 2n into the following n subsets {1, 2},{3, 4},, {2n-1, 2n}. Then if you take n+1 numbers from the set of the numbers from 1 to 2n, by the pigeonhole principle, at least two of them will be in the same subset. I.e., n + 1 numbers = pigeons n groupings of the 2n numbers = pigeonholes in the group of n +1 numbers at least two are consecutive, say k and k + 1. By (a), k and k+1 are relatively prime. There are at least two numbers that are relatively prime. (b) if n = 100 2n = 200 Can take the following set of 100 numbers from {1 200}: 2, 4, 6, 8, 10, 200 Since all the numbers are even, any pair of them has a common factor of 2. Therefore no pair is relatively prime.

4 6. Let S={ 3 + 4i i = 0,, 25}. If elements are randomly selected from S, how many elements must be selected from S to guarantee that at least two will sum to 110? Justify your answer and show all calculations. S = {3+4i i = 0,, 25} = {3, 7, 11, 15,, 103} S = 26 The following pairs sum to 110: (7,103) (11,99) (15,95) (19,91) (23,87) (27,83) 12 pairs (31,79) (35,75) (39,71) (43,67) (47,63) (51,59) The numbers 55 and 3 don t appear in any pair (since = 55 and there s only one 55 in the set and = 107 but the largest number in the set is 103) Claim If 15 numbers in the set are picked, then at least two of them will sum to 110. First of all, before we prove this claim, we argue that if only 14 are picked, you could end up with the numbers 3, 55 and the twelve first elements in each of the above pairs (namely, 7, 11, 15, etc) and since the first elements need the second elements to sum to 110, there won t be any two that sum to 110. So, no fewer than 15 numbers will guarantee that at least two sum to 110. Note: we are not claiming that there does not exist a collection of 14 numbers where at least two sum to 110. However, you are asked how many MUST be selected to GUARANTEE that at least two sum to 110. Since the selection is out of your control, you cannot assume that a selection of 14 will always guarantee that at least two sum to 110. If pick 15 elements, have 3 possibilities: (1) Case 1: 55 and 3 are both picked. Then 13 of the other numbers come from the twelve pairs.

5 Pigeons = numbers picked Pigeonholes = pairs 13 By Pigeonhole Principle, 2 12 must be in same pair (2) Case 2: Only one of 55 or 3 is picked. Then 14 of the other numbers come from the twelve pairs. 14 Again, by Pigeonhole Principle, 2 12 (3) Case 3: Neither 55 nor 3 is picked. Then 15 of the other numbers come from the twelve pairs. 15 By Pigeonhole Principle, 2 12 Since a selection of 14 or fewer numbers will not guarantee that at least two of the numbers sum to 110, by the above claim, 15 numbers must be picked to guarantee that at least two sum to 110. Note: since the claim proves that picking 15 numbers achieves the result, you cannot say that 16 (or any number higher than 15) must be picked to guarantee the result!