Question

Find factor(s) of $216$ which is divisible by both $3$ and $8?$

• A. $24$
• B. $48$
• C. $72$
• D. $216$

Answer 1

Answer: (A), (C), (D)

$216$

We know that 3 and 8 are relatively prime numbers.
Hence a number divisible by both 3 and 8 must be divisible by product of both 3 and 8.
$3\times 8=24$

$\therefore$ The required factor must be a multiple of $24.$

$\because 216=24\times 9\Rightarrow 216=24\times 3\times 3$

Hence, the factors of $216$ which are multiples of $24$ are $24\times 1,24\times 3\&24\times 9$ i.e., $24,72$ and $216.$ respectively.