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.