Problem 21: Show that the sum of two odd numbers is an even number.
Ans: an arbitrary odd number is represented as 2n + 1. So the sum of two odd numbers is
4n + 2 = 2(2n + 1), an even number..
Problem 22: What is the least common multiple (l.c.m) of 4, 8 and 12? What is the greatest common divisor (g.c.d) of 4, 8 and 12?
Ans: g.c.d. of a and b is denoted as (a, b). g.c.d. of a, b, and c is denoted as
(a, b, c) = ((a,b), c) = (a, (b,c)). So g.c.d of 4, 8, 12 =((4,8), 12) = 4.
l.c.m of a and b is denoted as [a,b]. l.c.m of a, b, and c is denoted as [[a,b],c] = [a,[b,c]]. So the l.c.m of 4, 8, 12 = [4,8],12] = 24. In general, the g.c.d and l.c.m of more than 3 numbers can be similarly determined.
l.c.m occurs frequently in the addition and subtraction of fractions when the least common denominator of the fractions are determined.
Problem 23: determine the g.c.d and l.c.m of 3, 6 and 9.
Ans: in general, the g.c.d and l.c.m of ma,mb and mc = m(a,b,c) and m[a,b,c] respectively. So (3,6,9) = 3(1,2,3) = 3. [3,6,9] = 3[1,2,3] = 18.
Problem 24: An integer > 1 is a prime number if its only divisors are itself and 1. All known prime numbers are odd numbers except one. Which even number is a prime number? .
Problem 25: Every integer > 1 can be written as a product of primes. Express 2500 and 64 as products of primes.
Ans: since √2500 = 50 and 50 is not a prime, only the primes below 50 need be examined. One method of prime factorization is to consecutively factorize the √ of the number of interest and then combine the factors as follows: 50 = 2(25) = 2(5)(5). So 2500 = 2(2)(5)(5)(5)(5) = 22. Simarly, 64 = 26
Problem 26: A prime divisor of the product of two or more integers divides at least one of the factors of the product. Which of the factors of 63 is a multiple of its prime divisor, 3?
Ans: 63 = 7(9). So 9 is a multiple of the prime divisor 3.
Problem 27: The factorization theorem states that every number can be represented uniquely as the product of prime numbers. Prove that there are infinitely many primes.
Ans: let C = a, b, c, ...,k be any collection of primes numbers . The sum of the product P of these primes and 1, S = P + 1, is either a prime or not a prime. S is a prime implies a new prime can be added to any collection of primes ad infinitum, therefore there are infinitely many primes. But suppose S is not a prime. This implies S must be divisible by some prime p that is not a member of C because if it is a member of C, it would divide P and P + 1; hence it will divide 1, which is their difference. This is impossible. Therefore, a new prime can always be found, given any set of primes. This proof is attributed to Euclid (Euclid's Elements Proposition 20, Book IX).
Problem 28: Determine the value of n for which the number 127 is a Mersenne prime .
Ans: If the number Mn = 2n - 1 is a prime, then it is a Mersenne prime. 127 = 27 - 1. So 127 is a Mersenne prime for n = 7.
Problem 29: suppose the binomial number N = a3 + b3 = 9, where a and b are integers. What is the value of a when b = 1?
Ans: the algebraic expansion of the binomal a3 + b3 = (a+b)(a2 - ab + b2) = 9. When b = 1, the expression becomes (a+1)(a2 -a + 1) = 9. Therefore, (a+1) is either 1, or 3 or 9. The equation is true only when (a+1) = 3. Therefore, a = 2.
Problem 30: all primes > 2 are odd numbers. Therefore, the minimum distance between any two consecutive primes is 2. Pairs of primes with this minimum distance between them are called prime twins. Suppose p1 and p2 are prime twins and p2 and p3 are prime twins. What is the sum of the even number between (p1, p2) and the even number between (p2, p3) if p2 = 19
Peter Oye Sagay