mod 10 = mod 10 = 49 mod 10 = 9.


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1 SOLUTIONS TO HOMEWORK 2  MATH 170, SUMMER SESSION I (2012) (1) (Exercise 11, Page 107) Which of the following is the correct UPC for Progresso minestrone soup? Show why the other numbers are not valid UPC s Note: UPC stands for Universal Product Code, which is basically the product bar codes we had been discussing in lectures. Solution : For the first one, without the check digit, mod 10 = mod 10 = 49 mod 10 = 9. Clearly the check digit, 1, adds up to 9 to give 0 mod 10. Hence, this is a UPC indeed. For the second one, without the check digit, mod 10 = mod 10 = 22 mod 10 = 2. But the check digit here is 2, and mod 10 = 4 0. So, this is not a UPC. For the third one, without the check digit, mod 10 = mod 10 = 5 mod 10. But the check digit is 3, and mod 10 = 8 0 So, this is not a UPC. (2) (Exercise 13, Page 107) The following is the UPC for Hellman s 8 oz. Real Mayonnaise. Find the missing digit
2 2 SOLUTIONS TO HOMEWORK 2  MATH 170, SUMMER SESSION I (2012) Solution : Let us call this missing digit x. Then, without the check digit, x mod 10 = x + 12 mod 10 = x mod 10 = 3 + 3x mod 10. Now, we must have that, 3 + 3x + 2 mod 10 = 0, i.e., 5 + 3x mod 10 = 0. This is possible only if x = 5 which gives us the missing digit. (3) A bank identification number is a 9 digit number that occurs in the lower left hand corner of bank checks. Let the digits be n 1, n 2, n 3, n 4, n 5, n 6, n 7, n 8, n 9. For a 9 digit number to be a valid bank identification number, it must satisfy 7n 1 + 3n 2 + 9n 3 + 7n 4 + 3n 5 + 9n 6 + 7n 7 + 3n 8 + 9n 9 0 mod 10. (Exercise 21, Page 108) Determine the check digits (i.e. last digits) of the following bank codes: Solution : Let us call the check digit c. In the first code, we need, c 0 mod 10. = c 0 mod 10. = 0 + 9c 0 mod 10. So, the digit c needed to make 9c 0 mod 10 is 0. So, in the first code, the check digit is 0. In the second code, we need, c 0 mod 10. = c 0 mod 10. = 6 + 9c 0 mod 10. Now, the digit c needed to make 6 + 9c 0 mod 10 is 6. So, in the second code, the check digit is 6. (4) Show that the following statements are true: (a) mod 23. Solution : By Fermat s Little Theorem, mod 23. That is, mod 23. We can now square both sides to get, mod 23. Multiplying both sides by 5 3, we get, mod 23. That is, mod 23. Now, 125 = = So, 125 mod 23 is 10. Therefore, 5 47 mod 23 = 10.
3 SOLUTIONS TO HOMEWORK 2  MATH 170, SUMMER SESSION I (2012) 3 Next, 6 17 = 102 = = and so, 6 17 mod 23 = 10. Thus, mod 23. (b) mod 13. Solution : By Fermat s Little Theorem, mod 13, that is, mod 13. Taking a power of 4, mod 13. So, multiplying by 2 3 = 8 on both sides, mod 13. That is, 2 51 mod 13 = 8. Next, mod 13 = mod 13 = (13+3) (26+7) mod 13 = 3 7 mod 13 = 21 mod 13 = 8. Thus, mod 13. (5) Carefully look at the following mathematical reasoning. Write down what is wrong with it. I believe that mod 17. This is because, Fermat s theorem says that if p is prime, a p 1 1 mod p, and since 17 is a prime with 17 1 = 16, we must have, mod 17. Taking a power of 2 on both sides of the congruence, we get, mod 17. After you have answered what is the mistake above, write down the correct number between 0 and 16 that is mod 17. Solution : Fermat s little theorem says that if p is prime and a is an integer NOT DIV ISIBLEBY p, then, a p 1 1 mod p. Although, 17 is prime here, 68 is indeed divisible by 17 (17 4 = 68), and so Fermat s Little Theorem does not apply here. The correct answer should be mod 17 = 0. This is because, as 17 divides 68, 17 will divide any power of 68, and so, 17 divides (6) Prove the following statements using the principle of induction: (a) (2n 1) = n 2, for all natural numbers n. Solution : The statement S(n) is (2n 1) = n 2. Step 1: We check if this is true for n = 1. The left hand side is just 1, and the right hand side is 1 2 and hence is 1. So, S(1) is true. Step 2: We assume S(m) is true. That is, we assume, (2m 1) = m 2. Step 3: Now we need to prove that S(m + 1) is true. That is, we need
4 4 SOLUTIONS TO HOMEWORK 2  MATH 170, SUMMER SESSION I (2012) to prove that (2(m + 1) 1) = (m + 1) 2. The left hand side is (2m+1) = (2m 1)+(2m+1) = m 2 +2m+1 (from Step 2). But, m 2 +2m+1 = (m+1) 2. Therefore, (2(m+1) 1) = (m+1) 2, which completes our induction steps. Hence, by induction, the statement (2n 1) = n 2 is true for all natural numbers n. (b) 3 2n n+1 is divisible by 7, for all natural numbers n. Solution : The statement S(n) is 3 2n n+1 is divisible by 7. Step 1: We check if this is true for n = = = = 7, which of course is divisible by 7. So, S(1) is true. Step 2: We assume S(m) is true. That is, we assume, 3 2m m+1 is divisible by 7. So, there is an integer k for which, 3 2m m+1 = 7k. Then, 2 m+1 = 7k 3 2m 1. Step 3: Now we need to prove that S(m + 1) is true. That is, we need to prove that 3 2(m+1) (m+1)+1 is divisible by 7. Now, 3 2(m+1) (m+1)+1 = 3 2m m+1 2 = 3 2m+1 + (7k 3 2m 1 ) 2, (from Step 2) = 14k + 3 2m m 1, = 14k + 3 2m m 1, = 14k + 3 2m 1 (3 2 2) = 14k + 3 2m 1 7, which, of course, is divisible by 7. This completes our induction steps. Hence, by induction, 3 2n n+1 is divisible by 7, for all natural numbers n. (c) Let F n denote the nth Fibonacci number (so, F 1 = 1, F 2 = 1, F 3 = 2, F 4 = 3 and so on). Show that F F F 2 n = F n F n+1, for all natural numbers n. Solution : The statement S(n) is F1 2 + F F n 2 = F n F n+1. Step 1: We check if this is true for n = 1. The left hand side F1 2, which is just 1, and the right hand side is F 1 F 2 = 1 1 = 1. So, S(1) is true.
5 SOLUTIONS TO HOMEWORK 2  MATH 170, SUMMER SESSION I (2012) 5 Step 2: We assume S(m) is true. That is, we assume, F F F 2 m = F m F m+1. Step 3: Now we need to prove that S(m + 1) is true. That is, we need to prove that F1 2 + F F m+1 2 = F m+1 F m+2. The left hand side is F1 2 + F F m 2 + Fm+1 2 = F m F m+1 + Fm+1 2 = F m+1(f m + F m+1 ). Now, because F n is the nth Fibonacci number, F m + F m+1 = F m+2. Therefore, F1 2 + F F m+1 2 = F m+1 F m+2, which completes our induction steps. Hence, by induction, the statement F F F 2 n = F n F n+1 is true for all natural numbers n. (7) Prove the following statements using the method of contradiction: (a) The negative of an irrational number is irrational. Solution : Let x be an irrational number. Assume that x is rational. So, we can write x as p q where p, q are integers with q 0. Thus, x = p q = x = p q = p q. Since p and q are still integers, with q 0, x is a rational number, which is a contradiction. Hence, the negative of an irrational number is irrational. (b) There are no even primes that are bigger than 2. Solution : Suppose there are even primes bigger than 2. Let p be one such prime. Since p is even, this means that we can write p = 2k for some natural number k > 1. But, this implies that both 2 and k divide p, which is a contradiction, since p is prime. Hence, there are no even primes that are bigger than 2. (c) 6 is irrational. Solution : Assume 6 is rational. Then we can write, 6 = p q, where p and q are integers that have no common factors, and q 0. Squaring both sides, 6 = p2 q 2, 6q 2 = p 2.
6 6 SOLUTIONS TO HOMEWORK 2  MATH 170, SUMMER SESSION I (2012) Thus, 6 divides p 2. This means that both 2 and 3 divide p 2. Now, from what we did in class, 2 divides p 2 means 2 divides p and 3 divides p 2 means 3 divides p. That is, both 2 and 3 divide p, which gives us that 6 divides p. So, we can write p = 6k, for some k. Then, plugging this in the previous equation we get, 6q 2 = (6k) 2 = 36k 2, = q 2 = 6k 2. So, 6 divides q 2. By the same reasoning as before, this means that 6 divides q. Therefore p and q have a common factor 6, which is a contradiction. Therefore, 6 is irrational. (d) is irrational. Solution : Assume is rational. Then we can write, = p q, where p and q are integers that have no common factors, and q 0. Squaring both sides, ( ( 2 + 3) = p2 q 2, = = p2 q 2, = 5q 2 + 2q 2 6 = p 2, = 6 = p2 5q 2 2q 2. The right hand side above is clearly a rational number, with the numerator and denominator both being integers and the denominator being nonzero. Therefore, the above expression shows that 6 is a rational number, which is a contradiction by what we proved in part (c) of this problem. (8) Let A and B be two sets. Draw Venn diagrams and shade the regions for the following set expressions:
7 SOLUTIONS TO HOMEWORK 2  MATH 170, SUMMER SESSION I (2012) 7 (a) (A B) c. Solution : The shaded area is the required region. (b) (A B) A. Solution : The area shaded in dark grey is the required region.
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