Question

A trade has 612 dettol soaps and 342 pear soaps. He packs them in boxes and each box contains exactly one type of soap. If every box contains the same number of soaps, then find the number of soaps in each box such that the number of boxes is the least.

Answer 1

Hint: This question talks about the number of soaps in each box can be packed so that number of boxes is the least ,so for this we have to find HCF (HIGHEST COMMON FACTOR) of both numbers and the HCF is our answer. HCF can be find by using two method
1. By using Euclid’s division algorithm
2. by using prime factorization method

Complete step-by-step answer:
We used HCF to find the answer because in the question it is given that the entire box contains only one kind of soap and the box must be full. As the name suggests the highest common factor is the largest number which divides both the given number. We considered the largest number because in the question it is given that we have to make as many less boxes as possible.
According to the question we have 612 dettol soaps and 342 pear soaps and we have to find the maximum number of soaps in each box.
To find the maximum number of soaps in each box we take HCF (HIGHEST COMMON FACTOR) of 612 and 342.
By using Euclid’s division algorithm, we have
$612 = 342 \times 1 + 270$
$342 = 270 \times 1 + 72$
$270 = 72 \times 3 + 54$
$72 = 54 \times 1 + 18$
$54 = 18 \times 3 + 0$
Here we notice that the remainder is zero, and the divisor at this stage is 18. Therefore HCF of 612 and 342 is 18.
So, the trader can pack 18 soaps per box.

Note: alternative method for this question
We can find HCF by prime factorization method where we have to write prime factor of \[612\] and $342$
So prime factor of $612$ is
$612 = 2 \times 2 \times 3 \times 3 \times 17$
And prime factor of $342$ is
$342 = 2 \times 3 \times 3 \times 19$
Now for HCF we have to collect common factor so we can see 2 ,3,3 are common factor
So HCF ($612,342$) = $2 \times 3 \times 3$
HCF ($612,342$) = $18$
So find the maximum number of soaps in each box = 18