Question

A sweet shopkeeper prepares 396 gulab jamuns and 342 rasgullas. He packs them into containers, each container consists of gulab jamun or rasgulla, but has an equal number of pieces he should put in each container so that numbers of containers are least.

Answer 1

Hint: First look at the definition of highest common factor, try to relate it to the question given. Now do prime factorization of both numbers. Calculate the highest common factor, this highest common factor is the required result.

Complete step-by-step answer:
Highest common factor: In mathematics the greatest common divisor of 2 or more numbers, which are not all zero, is the largest positive integer that divides each of the integers.
Process to find the highest common factor: Take the prime factorization of all the numbers and then try to find all the primes which are in common from all the numbers in the group, combine them to one and multiply. This product is nothing but the highest common factor.
Given condition in the question can be written in form of:
There are 396 gulab jamuns and 342 rasgullas equally packed. So, the highest common factor would be the least number of packs to get the result. For finding the highest common factor or greatest common divisor of the 2 numbers 342, 396 we need the prime factorization form of the two numbers 396, 342.
Prime factorization: In number theory, prime factorization is the decomposition of a composite number into a product of few prime numbers which are smaller than the original number. This process is carried out by dividing with prime number and finding quotient thus writing number as prime x quotient. Repeat the process for the quotient till you get 1 as the quotient.
Prime factorization of 342, by dividing with 2 we get,
$342=171\times 2$
By dividing the term 17 with 3 we can write it as,
$342=57\times 3\times 2$
By dividing the term 57 with 3, we can write it as,
$342=19\times 3\times 3\times 2$
By dividing the term 19 with 19 we can write it as,
$342=2\times 3\times 3\times 19\times 1$
We have to stop at this step as we got 1 as the quotient. Prime factorization of 396, by dividing with 2 we get,
$396=198\times 2$
By dividing the term 198 by 2, we can write it as,
$396=2\times 2\times 99$
By dividing the term 99 by ${{3}^{2}}$ we can write it as,
$396=2\times 2\times 3\times 3\times 11$
By dividing the term 11 by 11 we can write it as,
$396=2\times 2\times 3\times 3\times 11\times 1$
We have to stop here as we got the quotient as 1. The prime common are 2, 3, 3. So, by product of these
Highest common factor = $2\times 3\times 3=18$
So, there must be at least 18 boxes to satisfy the given condition.

Note: The idea of projecting the whole question into a single term, the highest common factor is very important. You must learn application of the highest common factor, least common multiple on to the real life incidents. While taking common if you get two 3’s as common, as two pairs of 3’s then 3 must be multiplied twice and do not even combine them.