Primes, HCF and LCM

Chapter 1 Study Notes

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1.3 Index Notation and Prime Factorisation

Index notation writes repeated multiplication short. In 53=1255^3 = 125, the base is 55 and the index is 33.

Index notation
an=a×a××an timesa^n = \underbrace{a \times a \times \cdots \times a}_{n \text{ times}}

Prime factorisation writes a composite as a product of primes. Every composite has exactly ONE such product however you reach it, and two methods do.

Split into any two factors, then keep splitting the composite ones until every branch ends on a prime. Any first split reaches the same answer.

Factor tree for 360
36049022910  3325\begin{array}{ccccccccc} & & & & 360 & & & & \\ & & & \swarrow & & \searrow & & & \\ & & 4 & & & & 90 & & \\ & \swarrow & & \searrow & & \swarrow & & \searrow & \\ \boxed{2} & & & \boxed{2} & 9 & & & 10 \\ & & & & \swarrow \; \searrow & & \swarrow & \searrow & \\ & & & \boxed{3} & & \boxed{3} & \boxed{2} & & \boxed{5} \end{array}

Divide by the smallest prime that goes in, repeatedly, until you reach 11.

Try it

A practice question on this appears here. Practise primes hcf lcm.

Repeated division (ladder method) for 360
23602180290345315551360=23×32×5\begin{array}{r|l} 2 & 360 \\ 2 & 180 \\ 2 & 90 \\ 3 & 45 \\ 3 & 15 \\ 5 & 5 \\ & 1 \end{array} \quad \Rightarrow \quad 360 = 2^3 \times 3^2 \times 5

Finding prime factorisation

Express 360360 as a product of its prime factors in index notation.

  1. Repeated division:
  2. 360÷2=180360 \div 2 = 180
  3. 180÷2=90180 \div 2 = 90
  4. 90÷2=4590 \div 2 = 45
  5. 45÷3=1545 \div 3 = 15
  6. 15÷3=515 \div 3 = 5
  7. 5÷5=15 \div 5 = 1
  8. 360=23×32×5\therefore 360 = 2^3 \times 3^2 \times 5

Try it

A practice question on this appears here. Practise primes hcf lcm.

Using prime factorisation to solve a problem

A rectangular floor is tiled with exactly 385385 identical square tiles. Both side lengths, in tiles, are greater than 11. Find every possible pair.

  1. 385=5×7×11prime factorisation385 = 5 \times 7 \times 11 \quad \scriptsize\textit{prime factorisation}
  2. Since each dimension >1> 1, we need two factors of 385385, both >1> 1:
  3. 385=5×77=7×55=11×35385 = 5 \times 77 = 7 \times 55 = 11 \times 35
  4. \therefore the possible dimensions are 5×775 \times 77, 7×557 \times 55, or 11×3511 \times 35 tiles.

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