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### Mopey & Zegro

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Date: Thu, 3 Nov 1994 12:17:18 -0700 (MST)
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?

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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
excellent.  Maybe the person left can carry 2 h's and 10 c's or
was great.

Ethan
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Associated Topics:
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