Question

Given that $(p+3)$ is the H.C.F 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 factors of 81 and 108 are:

81 = 3 $\times$ 3 $\times$ 3 $\times$ 3

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

The largest or greatest factor common to any two or more numbers is termed as H.C.F of given numbers.

$\therefore$ HCF of 81 and 108 = 3 $\times$ 3 $\times$ 3 = 27 --- (1)

But given, HCF of 81 and 108 = $(p+3)$ ------ (2)

Equating (1) and (2) we get,

27 = (p + 3)

$\Rightarrow$ p = 27 - 3 = 24