Mopey & Zegro

Date: Thu, 3 Nov 1994 12:17:18 -0700 (MST)
From: Laura Adams
Subject: Math

Boggs can carry two hartsels or three choloes. Mopey can carry three
hartsels or five choloes. Zegro can carry 7 hartsels or nine choloes.
They have to move 23 hartsels and 23 choloes out to the school 
playground. You will tell them what to carry you will have to carry 
whats left over. If each one makes 2 trips. What will be left over for 
you to carry?

Laura Adams (ladams@mtview.d20.co.edu)

Date: Thu, 3 Nov 1994 16:52:30 -0500
From: Phil Spector
Subject: Re: Math

Laura -

I think I have a solution.  Here's a hint:

Every time Mopey and Zegro carry objects, they can clearly carry 
more choloes than hartsels.  Here, it may be a good idea just to get 
as many objects carried as possible, regardless of what they are, in 
order to get as close as possible to carrying your total quota.  
However, Boggs is special, in that he can carry -relatively- more 
hartsels than choloes; you don't get the usual tradeoff of carrying 
2 less hartsels than choloes. Take advantage of Boggs' uniqueness 
and have him carry those hartsels!

See what kind of answers you get then.  The above plan should 
provide an answer, if the goal is for you to have to carry the least 
amount of objects.  In fact, I get two answers.  Do you think you 
can come up with a reason as to why one of them is a better answer 
for you?

Dr. Math a la Phil
__________

Date: Thu, 3 Nov 1994 19:56:04 -0500
From: Phil Spector
Subject: Laura's problem

Anybody (particularly Ethan/Sydney, who I heard were working on it) 
get a more precise answer to Laura's problem?  I think I got an answer 
(maximum you can get them to carry is 30 objects, with the best 
solution being having them carry 19 C and 11 H = (Zegro carrying 9 C 
and 7 H, Mopey carrying 5 C twice, and Boggs carrying 2 H twice, for 
19 C and 11 H.  This is better than the other solution, with 23 C and 7 H, 
since it seems as if H is the bulkiest (everyone can carry less of them 
than the others), so you want your friends to carry it as much as possible.

The above solution also assumes we're looking to carry the least 
-number- of total objects.  Note that I don't have a definite equation for 
the above math... it's a bit of trial and error.

Dr. Phil
__________

Date: Thu, 3 Nov 1994 20:02:15 -0500 (EST)
From: Dr. Ethan
Subject: Re: Laura's problem

Phil, that is exactly the way I did it and the answer I got. I think we 
need to know more about the goal of the problem to really give a 
helpful answer, although given the information given yours was 
excellent.  Maybe the person left can carry 2 h's and 10 c's or 
maybe they carry them evenly? So given what we had your answer 
was great.

Ethan