Question

Given that $(p+3)$ is the highest common factor of 81 and 108, find the value of $p$.

• A. 6
• B. 12
• C. 9
• D. 24

Answer 1

The correct option is D

24

Prime factorisation of 81 and 108:

81 = 3 $\times$ 3 $\times$ 3 $\times$ 3
108 = 2 $\times$ 2 $\times$ 3 $\times$ 3 $\times$ 3

$\therefore$ The highest common factors is 3 $\times$ 3 $\times$ 3 = 27

Given, HCF of 81 and 108 is $(p+3)$.
$\therefore p+3=27$
$\Rightarrow p=24$