Problem (1):
I have some pencils and some jars.
If I put 4 pencils into each jar I will have one jar left over.
If I put 3 pencils into each jar I will have one pencil left over.
How many pencils and how many jars?

Problem (2):
Again I have some pencils and some jars.
If I put 9 pencils into each jar I will have two jars left over.
If I put 6 pencils into each jar I will have three pencils left over.
How many pencils and how many jars?

(In reply to about it: by Gamer)

Or from a more formal mathematical presentation:
let j = #jars
let p = #pencils
then for problem 1 have 4(j-1) = p and 3j+1 = p. Solving for j have 4j-4 = 3j+1, or j=5. Then solving for p have 3(5)+1 = p = 16. Thus the answer shown above, 5 jars and 16 pencils.

for problem 2 have 9(j-2) = p and 6j+3 = p, so 9j-18=6j+3, so 3j=21, j=7. Then solving for p have 9(7-2) = p, 9(5) = p, so p=45. Thus the answer for problem 2 is seven jars and forty-five pencils.

I agree that it would really be a waste of time to use a Supercomputer to solve this basic algebra problem.

Posted by Jennifer on 2003-12-17 00:17:55