Question

anu has 138 pencils and anuj has 207 pencils. if they pack their pencils in identical boxes, each containing an equal number of pencils, then find the greatest number of pencils that can be put in each box, without any left over.

Answer 1

This can be solved using the stars and bars method.[1]

First, since everyone needs at least 1 chocolate, we can start by handing out 1 chocolate to every student.

Now we have 3 more chocolates left to give to the 4 students.

Since nobody can have more than 3 chocolates, we cannot give all 3 chocolates to the same student.

Therefore, this is the same thing as the number of ways of giving 3 identical chocolates to 4 students such that no student gets more than 2 chocolates.

Using the stars and bars algorithm, the ways of distributing n identical objects to k different groups is:

(n+k−1n)(n+k−1n)

So for this problem, we can start with the ways of giving 3 identical objects to 4 different people.

(3+4−13)=(63)=20(3+4−13)=(63)=20

However, this count includes counts where all 3 chocolates are given to the same student, which is not desired. Therefore, we must subtract how many ways there are to give all the chocolates to a single student. Since there are 4 students there are 4 ways of doing that. Therefore, after subtracting, the number of ways is:

20−4=1620−4=16

Answer 2

Prime Factorisation of 138 and 207:

138 = 2 x 3 x 23

207 = 3² • 23

Find HCF:

HCF = 3 x 23 = 69

Answer; the greatest number of pencils that can put in each box is 69