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The PCTM Puzzle of the Week - Solutions

We are assuming quite a bit for this problem, so 
there are many solutions, if one looks deeply enough.

First, the solution that is probably wanted is:

Since the total number of nuggets needed is 8+5+5+4 = 22,
we must have at least 22 nuggets.  6 and 9 are multiples of 
3, so any combination of those must also be a multiple of 
three -- the 
smallest above 22 is 24, which is able to be found 
using: 
6+6+6+6=24
OR
9+9+6=24.

We would not use the 20 in this case, as the 
smallest number of
nuggets if we use 20 would be 20+6=26, which is 
larger than our
optimal answer of 24. 

YET, the question states that Mr. McKluck COULD eat 
8 -- he is
not necessarily going to get 8, so the optimal 
answer could, in
fact be 1+5+5+4=15=6+9. :*}

I don't know the prices for nuggets, so I'll skip 
that, and move
to extra question number 2

Answer: 43

Proof:
(For my proof, remember that X (mod 3) is equal to 
the 
remainder when X is divided by 3 (so it can have 
values of 0,1,2))

As I said before, 6 and 9 are multiples of 3, and as 
can easily 
be seen, any multiple of 3, greater than 6 can be 
made with a 
combination of 6's and 9's.  This means that the 
highest number,
equivalent to 0 (mod 3), that can not be represented 
as a sum of
nugget groups, is 3.

For numbers not equal to 0 (mod 3), we must have 
at least
one 20 (which is equiv. to 2 (mod 3))

This means that any number equiv to 2 (mod 3), 
greater than
20+6=26 can be represented in terms of one 20 plus 
some number
of 6's and 9's, so the highest number equivalent to 
2 (mod 3) 
that can not be represented as a sum of nugget 
groups is 23.

Lastly, for the only case left (a number is 
equivalent to 
1 (mod 3)), we must have two 20's, so any number 
equivalent to 
1 (mod 3) greater than 40+6=46 can be represented 
in terms of 
two 20's and some combination of 6's and 9's, so 
the highest
number equivalent to 1 (mod 3) that can not be 
represented as
a sum of nugget groups is 43.

Since 43>23>3, 43 is the largest number of nuggets 
such that it
can not be represented as a sum of any 
combinations of 6's, 9's
or 20's.

This may be a bit too involved, but to prove it, one 
must be 
rigorous. (This isn't even terribly rigorous)

Bye!
--Christophe

He ordered either 4 6pc. nuggets, or 2 9pc. and 1 
6pc.  I have to check
the price of McNuggets when I go to McDonalds 
next, and I'm not totally 
clear on the Third part.  I've tried making a 
sequence out of it, but
I don't see one. (Trying to get the HIGHEST possible 
figure).

I used try and check for this problem. I added the 
numbers given, 8, 5, 5, and 4. Then I tried adding 
different sizes of chicken mcnuggets until I found 
two compatible numbers. I totally eliminated the 
20 size mcnuggets since I had to get 22. My final 
answer was 2 packs of 9 and 1 pack of 6. You will 
have 2 extra nuggets.

6+6+6+6=24
6+9+9=24
If you buy 24 Mcnuggets you will have 2 left over.

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