SOLUTION: There is a specific number of pens available for three students. The first student takes one third of the pens, the second student also takes one third of the pens and the third st

Algebra ->  Divisibility and Prime Numbers -> SOLUTION: There is a specific number of pens available for three students. The first student takes one third of the pens, the second student also takes one third of the pens and the third st      Log On


   

Question 769969: There is a specific number of pens available for three students. The first student takes one third of the pens, the second student also takes one third of the pens and the third student takes one third of the remaining third of the pens. The students distribute the remaining pens evenly among each other. What was the initial total number of pens?
Answer by ramkikk66(644) About Me  (Show Source):
You can put this solution on YOUR website!
Let the total number of pens be x
1st student takes x/3
2nd student also takes x/3
Remaining is (x - x/3 - x/3) = x/3
3rd student takes 1/3rd of this, i.e. x/9
Number of pens taken = x/3 + x/3 + x/9 = 7*x/9
Remaining = x - 7*x/9 = 2*x/9
This is divided evenly amongst the 3, so each gets 2*x/(9*3) = 2*x/27
For this to be a whole number, x has to be a multiple of 27.
The smallest value of x that satisfies this is, of course, 27.
So the smallest number of pens that satisfies the conditions is 27.
Check:
1st person gets 9
2nd student gets 9
Remaining 9
3rd students takes 1/3rd = 3
Total taken = 9 + 9 + 3 = 21
Remaining 6
Divided equally among the 3, each gets 2.
Note: 27 is the *least* number that satisfies the conditions.
Take any multiple of 27, those work for the conditions as well. e.g. 54, 81, 108. So actually this problem has infinite number of solutions.
:)