Homework for Oct 13th -Ancient Egyptian 'algebra': The method of 'false position'

 Here is the new question : 

Sam and Jack had 80  erasers in total. After Sam giving  1/3 of his erasers and Jack, Sam and Jack had an equal number of erasers. How many erasers did Sam have in the beginning?

Using modern Method: Let the numbers of the erasers in the beginning=X,

then we have X-(X/3)=40,

we can simplify it 2X/3=40,

So, we get X=60

Using the Egyptian 'false position' method,

we have to choose the number  divisible by 3, let X=9,

then, we have 9-(9/3)=6, but 40 is not the multiple of 6, so try again, 

let X=12, then 12-(12/3)=8, but we need 40, which is 5 times as big as 12.  

we have X=12*5=60, we check 60 with the equation, 60-(60/3)=40

Therefore, in the beginning, Sam had 60 erasers.