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?

Let the respective numbers of pencils and jars be x and y.

For part (a); we obtain:
x = 4(y-1) = 3y + 1.
Solving, we get, x = 16, y = 5; so that the respective numbers of pencils and jars are 16 and 5.

For part (b); we obtain:
x = 9(y-2) = 6y + 3.
Solving, we get, x = 45, y = 7; so that the respective numbers of pencils and jars are 45 and 7.

This puzzle confused me at first.
Nice problem.

Edited on March 16, 2007, 10:41 am

Posted by K Sengupta on 2007-03-16 10:39:13