In Levikland, there are coins worth 1, 2, 5, 10, 20, 50 and 100 perplexii. A has twice as much money as B, who has twice as much as C, who has twice as much as D. How can this be, if everybody has two coins?

The answer is 120-60-30-15.  Person A has 100+20 perplexii, B has 50+10 perplexii, C has 20+10 perplexi, and D has 10+5 perplexi.

The way I solved it was just by making a little table of all the possible 2 coin sums. I started looking at the smaller numbers and seeing if their doubles existed (2, 4, op there’s no 8… 3, 6, 12, op there’s no 24). Then I started looking the other way at the larger numbers (200, 100, op no 50… 150, op no 75… 120, 60, 30, 15!!!).

However, I only tell you that out of honesty. There’s probably a more systematic way of solving it.

Person A’s sum must be divisible by 8, otherwise you won’t have integers for everyone else’s sum. Well, if it is still ok to at least have my table as reference, I can make another table that is my first table divided by 8. Only the whole number results are possible cases for A’s sum. The only whole number entries were when A’s sum was 40, 120, and 200.

So this is only 3 cases to check… which I think is more systematic than my random "let’s see if this works – no… let’s see if this works – no…" approach.

It’s easy to see that you can’t solve for D when A has 40 or 200. You can’t make D have 5 or 25 with only two coins, so 120-60-30-15 must be it.

 

Is this a legal approach? It’s at least better than what I really did =)

Posted by nikki on 2004-11-19 14:27:38