Mathematics 222 (A1) Midterm Examination May 24, 2002

Transcription

1 Mathematics (A) Midterm Examination May 4, 00 Department of Mathematical and Statistical Sciences University of Alberta Instructor: I. E. Leonard Time: 70 Minutes. There were 5 children in the Emergency Room and between them they had stuck a total of 76 beans up their noses. Show that there must be three children with a combined total of 45 or more beans up their noses. Solution: Suppose there do not exist 3 children with a combined total of 45 or more beans up their noses. If we label the children A, B, C, D, and E and a, b, c, d, and e are the number of beans in each child s nose, respectively, then we have x + y + z 44 ( ) 5 for any triple x, y, z {a, b, c, d, e}. There are = 0 ways to choose the triplet x, y, z, so that we 3 get 0 inequalities like this. Namely, a + b + c 44 a + b + d 44 a + b + e 44 a + c + d 44 a + c + e 44 a + d + e 44 b + c + d 44 b + c + e 44 b + d + e 44 c + d + e 44 Each of the numbers a, b, c, d, e appears in exactly of these 0 inequalities. ( ) 4 = 6 of the expressions on the left hand side If we add these 0 inequalities, on the left hand side we get 6 (a + b + c + d + e), while on the right hand side we get 0 44, that is, 6 (a + b + c + d + e) 0 44, so that 456 = = 440. Which is obviously a contradiction. Therefore, our original assumption must have been incorrect, and there must be three children with a combined total of 45 or more beans up their noses.

2 . Dr. Ecco concluded that the aliens from whom the message was received also had an alphabet consisting of 6 distinct characters. Another moment s thought convinced him that the code was a linear code with encoding function E(x) 5x (mod 6). Your mission, should you decide to accept it, is to help Dr. Ecco find the decoding function D(x), that is, find an integer a such that D(x) ax (mod 6), and decode the following message from the aliens: IH FUAN, KADD JSIU For your convenience, the correspondence between the letters of the alphabet and the integers 0 5 is given below. A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Solution: Since 5 05 (mod 6), then the decoding function is D(x) x (mod 6), for x = 0,,..., 5. Decoding the message, we have I 8 D(8) 8 68 (mod 6) M H 7 D(7) (mod 6) 7 R F 5 D(5) 5 05 (mod 6) B U 0 D(0) (mod 6) 4 E A 0 D(0) 0 0 (mod 6) 0 A N 3 D(3) (mod 6) 3 N K 0 D(0) 0 0 (mod 6) C A 0 D(0) 0 0 (mod 6) 0 A D 3 D(3) 3 63 (mod 6) L D 3 D(3) 3 63 (mod 6) L J 9 D(9) (mod 6) 7 H S 8 D(8) (mod 6) 4 O I 8 D(8) 8 68 (mod 6) M U 0 D(0) (mod 6) 4 E The message reads: MR BEAN, CALL HOME

3 3. The telephone numbers in a town run from to A common error in dialing on a standard keypad is to punch in a digit vertically adjacent to the intended one. So, on a standard dialing keypad, 6 could be erroneously entered as 3 or 9 (but not as, 5 or 8) Assuming that no other kinds of errors are made, how should a sixth digit be added to each telephone number so that no wrong numbers will be reached because of a dialing error? Solution: Again, there are many acceptable solutions. Given the number abcde, one solution is to choose the sixth digit f so that a + b + c + d + e + f 0 (mod 0), in other words, so that a + b + c + d + e + f is divisible by 0. Note that if an error is made then the digits that get switched differ by 3 or 8. Suppose that in abcdef a digit x gets changed to x, for example, suppose that x = b. If the error is not detected, then a + x + c + d + e + f 0 (mod 0) and a + x + c + d + e + f 0 (mod 0), and so x x 0 (mod 0) However, x x 0 (mod 0) if and only if x x is a multiple of 0, and since x and x are between 0 and 9, then 9 x x 9, but the only multiple of 0 between 9 and 9 is 0 = 0 0. Therefore, if the error is not detected, then x = x, or equivalently, if x and x are switched and are different, then the proposed scheme will always catch the error. 4. A published book is identified by a ten-digit number known as its International Standard Book Number or ISBN. These digits denote the language, the publisher and other data concerning the book, except for the last one, which is introduced as a check-digit. If the first nine digits are a b c d e f g h i, then the tenth digit j is chosen so that 0a + 9b + 8c + 7d + 6e + 5f + 4g + 3h + i + j 0 (mod ). If j has to be ten, an X is used instead. In the following ISBN find the digit that goes in the square Solution: We choose the digit j so that j 0 (mod ), or, j 0 (mod ), that is, j 36 6 (mod ). Therefore, the digit that goes in the square is 6. 3

4 5. The two-out-of-five code consists of all possible binary words of length 5 containing exactly two s. (a) How many code words does the code contain? (b) What is the minimum Hamming distance between code words? (c) How many errors can the code detect or correct? Solution: (a) The number of distinct code words is ( ) 5 = 5 4 = 0, and we can list them as follows (b) Let a and b be two distinct code words. If the number of positions where the two s overlap is 0, for example, a = and b = 0 0 0, then the Hamming distance is d(a, b) = 4. If the number of positions where the two s overlap is, for example, a = and b = 0 0 0, then the Hamming distance is d(a, b) =. Since these are the only two possibilities, then the minimum Hamming distance between any two code words in the two-out-of-five code is. (c) Since the minimum Hamming distance between any two code words is, this code can detect up to one error, but cannot correct any errors. 6. (a) Find a closed form expression for n. (b) Make a conjecture about the terms of the following sequence, and prove your conjecture. +, , , ,... Solution: (a) We have and adding, we get S n = n + n S n = n + n S n = (+n)+(+n )+ +(n +)+(n+) = (n+)+(n+)+ +(n+)+(n+) = n(n+), so that S n = n(n + ). 4

5 (b) Since each of the terms shown above is equal to 3, if we let a n = n n + n n, for n, then it appears that a n = 3 for all n. In order to see that this is indeed the case, we use the result from part (a) to write for n. a n = n(n + ) n(n + ) n(n ) = n + 4n + (n ) = n + 3n + 3 = n + 3(n + ) = 3 5