Question

Find the HCF of 153, 117 and 405.

• A. 9
• B. 12
• C. 6
• D. 2

Answer 1

Answer:

A. 9

Let's find the HCF (Highest Common Factor) of (153), (117), and (405) step by step.

Factor 153:

$153 \div 3 = 51 \quad (\text{since } 1+5+3 = 9, \text{divisible by 3})$
$51 \div 3 = 17$
$17 \text{ is prime}$

So, $(153 = 3^2 \times 17)$

Factor 117:
$117 \div 3 = 39 \quad (\text{since } 1+1+7 = 9)$
$39 \div 3 = 13$
$13 \text{ is prime}$

So, $(117 = 3^2 \times 13)$

Factor 405:

$405 \div 3 = 135$
$135 \div 3 = 45$
$45 \div 3 = 15$
$15 \div 3 = 5$

So, $(405 = 3^4 \times 5)$

Find common factors

$(153 = 3^2 \times 17)$

$(117 = 3^2 \times 13)$

$(405 = 3^4 \times 5)$

The common prime factor is (3), and the smallest power is $(3^2 = 9)$.