State the Fundamental theorem of arithmetic for positive whole numbers greater than \(1\).

Use the Fundamental theorem of arithmetic, applied to \(5577\) and \(99\,099\), to calculate \(\gcd (5577,{\text{ }}99\,099)\) and \({\text{lcm}}(5577,{\text{ }}99\,099)\), expressing each of your answers as a product of prime numbers.

Prove that \(\gcd (n,{\text{ }}m) \times {\text{lcm}}(n,{\text{ }}m) = n \times m\) for all \(n,{\text{ }}m \in {\mathbb{Z}^ + }\).

every positive integer, greater than \(1\), is either prime or can be expressed uniquely as a product of primes     A1A1

Note:     Award A1 for “product of primes” and A1 for “uniquely”.

[2 marks]

\(5577 = 3 \times 11 \times {13^2}{\text{ and }}99099 = {3^2} \times 7 \times {11^2} \times 13\)     M1

\(\gcd (5577,{\text{ }}99099) = 3 \times 11 \times 13,{\text{ lcm}}(5577,{\text{ }}99099) = {3^2} \times 7 \times {11^2} \times {13^2}\)     A1A1

[3 marks]

METHOD 1

\(n = p_1^{{k_1}}p_2^{{k_2}} \ldots p_r^{{k_r}}\) and \(m = p_1^{{j_1}}p_2^{{j_2}} \ldots p_r^{{j_r}}\)     M1

employing all the prime factors of \(n\) and \(m\), and inserting zero exponents if necessary     R1

define \({g_i} = \min ({k_i},{\text{ }}{j_i})\) and \({h_i} = \max ({k_i},{\text{ }}{j_i})\) for \(i = 1 \ldots r\)     (M1)

\(\gcd (n,{\text{ }}m) = p_1^{{g_1}}p_2^{{g_2}} \ldots p_r^{{g_r}}\) and \({\text{lcm}}(n,{\text{ }}m) = p_1^{{h_1}}p_2^{{h_2}} \ldots p_r^{{h_r}}\)     A1A1

noting that \({g_i} + {h_i} = {k_i} + {j_i}\) for \(i = 1 \ldots r\)     R1

\(\gcd (n,{\text{ }}m) \times {\text{lcm}}(n,{\text{ }}m) = n \times m\) for all \(n,{\text{ }}m \in {\mathbb{Z}^ + }\)AG

METHOD 2

let \(m\) and \(n\) be expressed as a product of primes where \(m = ab\) and \(n = ac\)     M1

\(a\) denotes the factors that are common and \(b,{\text{ }}c\) are the disjoint factors that are not common     R1

\(\gcd (n,{\text{ }}m) = a\)     A1

\({\text{lcm}}(n,{\text{ }}m) = \gcd (n,{\text{ }}m)bc\)     A1

\(\gcd (n,{\text{ }}m) \times {\text{lcm}}(n,{\text{ }}m) = a \times (abc)\)     M1

\( = ab \times ac\) and \(m = ab\) and \(n = ac\) so     R1

\(\gcd (n,{\text{ }}m) \times 1{\text{ cm}}(n,{\text{ }}m) = n \times m\) for all \(n,{\text{ }}m \in {\mathbb{Z}^ + }\)     AG

[6 marks]

Total [11 marks]

In part (a), most candidates omitted the 'uniquely' in their definition of the fundamental theorem of arithmetic. A few candidates defined what a prime number is.

In part (b), a substantial number of candidates used the Euclidean algorithm rather than the fundamental theorem of arithmetic to calculate \(\gcd (5577,{\text{ }}99099)\) and \({\text{lcm}}(5577,{\text{ }}99099)\).

In part (c), a standard proof that has appeared in previous examination papers, was answered successfully by candidates who were well prepared.