Number Theory Problem
I am stuck on this tutorial problem;
A number theory course is given in 3 different lecture theatres - ThA (9 seats per row), ThB (10 seats per row) and ThC (11 seats per row).
The students always fill the back rows first, so that the backmost rows are completely full, the frontmost rows are completely empty and at most 1 row is partially full.
The frontmost row has 5 students in ThA, 1 student in ThB and 10 students in ThC.
How many students are there? Assume that all students attended all lectures. You need to make a reasonable assumption about the size of the lecture theatres.
So you are looking for the smallest natural number that can be written as 9a+5,10b+1 and 11c+10 where a,b,c are also natural numbers (or perhaps the second if it turned out that that number meant one of the lecture theatres was implausibly small) . The 10* table+1 is easy to do in your head, and the remainder when you divide by 9 will increase by 1 each with each multiple (until you get to 9*10). So you'll quickly find that 41 is the first number that leaves a remainder of 5 when divided by 9 and 1 when divided by 10. Now keep adding 90 until you get to a number that leaves a remainder of 10 when you divide it by 11.
Thanks heaps buddy, much appreciated (Wink)
The Chinese Remainder Theorem seems to be a natural tool for solution of this problem.
Originally Posted by Banana1
Chinese remainder theorem - Wikipedia, the free encyclopedia