1.3Index Notation and Prime Factorisation

Index notation writes repeated multiplication short. In $5^3 = 125$ , the base is $5$ and the index is $3$ .

Index notation

\displaystyle a^n = \underbrace{a \times a \times \cdots \times a}_{n \text{ times}}

Now you try

Write $18 \times 18 \times 18$ in index notation.

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18 \times 18 \times 18 \quad \scriptsize\textit{18 appears 3 times}

= 18^{3}

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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. Each number splits into two, and the boxed leaves are the primes.

Divide by the smallest prime that goes in, repeatedly, until you reach $1$ .

Repeated division (ladder method) for 360

\displaystyle \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 $\displaystyle 360$ as a product of its prime factors in index notation.

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Repeated division:

\displaystyle 360 \div 2

\displaystyle = 180

\displaystyle 180 \div 2

\displaystyle = 90

\displaystyle 90 \div 2

\displaystyle = 45

\displaystyle 45 \div 3

\displaystyle = 15

\displaystyle 15 \div 3

\displaystyle = 5

\displaystyle 5 \div 5

\displaystyle = 1

\displaystyle \therefore 360

\displaystyle = 2^3 \times 3^2 \times 5

Now you try

Express $116$ as a product of its prime factors in index notation.

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\displaystyle 116 \div 2

\displaystyle = 58 \quad \scriptsize\textit{repeated division}

\displaystyle 58 \div 2

\displaystyle = 29

\displaystyle 29 \div 29

\displaystyle = 1

\displaystyle 116

\displaystyle = 2 \times 2 \times 29

\displaystyle = 2^{2} \times 29 \quad \scriptsize\textit{index notation}

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Using prime factorisation to solve a problem

A rectangular floor is tiled with exactly $\displaystyle 385$ identical square tiles. Both side lengths, in tiles, are greater than $\displaystyle 1$ . Find every possible pair.

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385 = 5 \times 7 \times 11 \quad \scriptsize\textit{prime factorisation}

Since each dimension

> 1

, we need two factors of

385

, both

> 1

\displaystyle 385

\displaystyle = 5 \times 77

\displaystyle = 7 \times 55

\displaystyle = 11 \times 35

\therefore

the possible dimensions are

5 \times 77

7 \times 55

, or

11 \times 35

tiles.