Advancing a Millennium Problem

Outline Mathematics
Algebra, Word Problems

Find all solutions in positive integers to 2015=20·x+15·y.

Solution

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Copyright © 1996-2017 Alexander Bogomolny

Find all solutions in positive integers to 2015=20·x+15·y.

Solution

The problem is a modification of Problem #78 from David Singmaster's book Problems for Metagrobologists:

When the Millennium was approaching and the Dome and the Jubilee Line were still in the state of chaos, it was clear that we needed the extra year available as the next Millennium really didn't start until the beginning of 2001. This has very little to do with our problem except that it involves dates! Reading an 18C book, I was inspired to ask how many solutions are there to 19X+99Y=1999 with positive integers X an Y? After doing the whole thing out, I realize that I knew one solution, namely 19·100+99·1=1999 and one easily see there is only one other solution 19·1+99·20=1999. So I wondered about using 19 and 99 to make 2000. That is, how many solutions are there to 19X+99Y=2000 with positive integers X an Y?

I modified the problem and adapted David's method for solving his problem.

If we relaxed the requirements of the problem for a moment, it should make it easier to solve it, what do you think? Right,Right,Wrong,Don't know. Let's see. Is it easier to find a solution to the equation 2015=20·x+15·y if x and y are not required to be positive? Neither can be zero, but negative? It is easy to verify that 20-15=2015/403,400,401,402,403. This translates into 20·403+15·(-403)=2015. 2015 is now represented as the sum of two terms so that if we add to one and subtract from the other the same amount, the identity will remain in place. With this in mind, we wish to increase,decrease,increase the second term as to make it positive, without taking too much,too much,too little,anything at all from the first term. This quantity must be divisible by both 20 and 15, so the smallest it can be is 60,40,45,50,60.

What we are looking for then are numbers N such that 403-3N and -403+4N are positive,positive,negative. Indeed, then

20·(403-3N) + 15·(-403+4N) = 20·403 + 15·(-403).

We thus require 403-3N ≥ 0 and 403-4N ≤ 0. The two inequalities are satisfied by 403/4 ≤ N ≤ 403/3 so that any N in the range 100 ≤ N ≤ 134 will solve our problem. Wait, is this false,true,false? Well, the right answer is 100 < N ≤ 134 - a slight difference.

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Copyright © 1996-2017 Alexander Bogomolny

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