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Apple Dealers: CR4 Newsletter Challenge (November 2009)

Posted November 01, 2009 5:01 PM

This month's Challenge Question:

The three apple farmers (Mike, Juliet, and Bob) finished their harvest and reported yields of 314,827 apples, 1,199,533 apples, and 683,786 apples, respectively. Mike mentioned the number of apples he'd have left over if he divided his harvest equally among all the apple dealers in the area. Juliet indicated she would have bought the extra apples so she could divide her supply equally among the dealers, with Bob indicating the same thing. How many apple dealers are there?

The Answer will be posted right here on CR4 on December 2nd. Can't wait that long? Well, check out these weekly challenges from CR4:

Centrifuge Training: CR4 Challenge (11/24/09)

Rocket Science: CR4 Challenge (11/17/09)

Rescue Plane: CR4 Challenge (11/10/09)

Chain Links: CR4 Challenge (10/27/09)

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Guru
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#1

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/01/2009 7:36 PM
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#2

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/01/2009 11:28 PM

131 apple dealers.

mod (314,827 / 131) = 34

So Mike has 34 apples left over after distributing evenly to 131 dealers.

mod ((1,199,533 + 34) / 131) = 0

Juliet can buy the extra 34 apples from Mike and evenly distribute to 131 dealers.

mod ((683,786 + 34) / 131) = 0

Bob can also buy the extra 34 apples from Mike and evenly distribute to 131 dealers.

There might be a higher number of dealers, but this is the lowest number where this occurs.

Reply Good Answer (Score 3)
Anonymous Poster
#21
In reply to #2

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 10:22 AM

How did you know to divide the harvest yields by 131?

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Participant

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#40
In reply to #2

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/10/2009 6:16 PM

I know the maths is accurate, but 131 dealers is a lot in a local area... in today's world with centralized marketing, my vote goes for the simplest solution. There is only 1 dealer and the number of left-over apples Mike mentioned was zero. Who would buy apples from other growers to ensure 131 dealers all got the same number of apples anyway! Farmers have a dry sense of humor...when it comes to marketing monopolies.

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Guru

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#42
In reply to #40

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/11/2009 12:04 AM

what about a small city with a population of 131?

and everyone is his own dealer?

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Anonymous Poster
#3

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/02/2009 12:11 AM

The question answers itself "The three apple farmers," implying that there are only these three apple farmers.

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Commentator

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#4

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/02/2009 4:59 AM

Hi All

Wrote a little procedure

procedure TForm1.Button1Click(Sender: TObject);

type
apples = record
amount:integer;
left:integer;
end;
var
mike,juliet,bob:apples;
dealers,counter:integer;

begin
memo1.Lines.Clear;
mike.amount:=314827;
juliet.amount:=1199533;
bob.amount:=683786;
for dealers:=2 to 315000 do
begin
mike.left:=mike.amount mod dealers;
juliet.left:=(juliet.amount+mike.left) mod dealers;
bob.left:=(bob.amount+mike.left) mod dealers;
if (juliet.left=0) and (bob.left=0)
then memo1.Lines.add('Mike has left '+inttostr(Mike.left)+
' ,dealers = '+inttostr(dealers));
end;
end;

There are 131 dealers with Mike having 34 apples left

I ran the program up to 315 000 dealers. There are no other number of dealers.

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Commentator

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#5
In reply to #4

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/02/2009 6:54 AM

Or an exact solution as follows. == stands for congruent

m=314827==x mod y // y is the number of dealers

j=1199533 + x == 0 mod y

b=683786 + x == 0 mod y

So 2m + j + b = 2512973 + 2x == 2x mod y

So 2512973 == 0 mod y

2512973 has 131 and 19183 as prime factors. So we have

2512973 == 0 mod 131 or 19183

So there are 131 dealers and that would leave Mike with 34 apples left

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Anonymous Poster
#6

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/02/2009 8:58 AM

Why do they put quizzes answered within a few hours with no controversy?

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#7
In reply to #6

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/02/2009 9:13 AM

I think a GA for this challenge should be one that does not use a computer based solution. There must be a way to use logic based on factors or other features of the three apple production numbers to find the solution. We can use an Excel spreadsheet can find the answer but that is just an answer. What we should be striving for is a solution that requires only a sheet of paper, a pencil, and a sharp mind.

Thanks,

Jim

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7
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#8
In reply to #7

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/02/2009 9:51 AM

The requirement is that the number_of_dealers is a common factor of Mike's+Juliet's and Mike's+Bob's.
I.e. we want the common factors of 314,827+1,199,533 and of 314,827+683,786.
They factorise respectively as 23x5x172x131 and 32x7x112x131.
As 131 is the only common factor that must be the number of dealers.

Now for the bidding war between Juliet and Bob

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#9
In reply to #8

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/02/2009 11:58 AM

GA by any definition.

Thanks,

Jim

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Guru

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#10
In reply to #9

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/02/2009 1:15 PM

On the other hand, if the farmers want to sell all dealers the same number (not weight??) of apples, perhaps they would improve their prices by saying they would only support 128 dealers this year, and the dealers better make a higher offer if they want any apples. Then they could transfer a few apples between themselves, and sell all but two (or 126 etc. if they want to keep some of the saleable ones).

P.S. Pencil very sharp...

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#17
In reply to #9

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 5:45 AM

Agreed, well done Fyz!

It took me a minute or 2 to see that Mike dividing among the dealers and giving remainder to Juliet, who then has a whole multiple of dealers, is the same as Mike's + Juliet's total being a whole multiple of dealers.

Cheers........Codey

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#12
In reply to #6

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 1:54 AM

Could you solve the problem? Fact is that you could not. Thats why you hide your identity and prefered to remain as guest. Then why underestimate those, who solved the problem.

I accept, i could not solve it.

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Anonymous Poster
#14
In reply to #12

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 4:20 AM

Could be that guest did solve the problem. 50 years ago it was a standard for 15-year-olds learning about prime decomposition.

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Anonymous Poster
#20
In reply to #12

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 9:03 AM

Unfortunately I did, and more unfortunately, when I tajke a challenge I don't look at the answers already made. And even more I logged in late, and saw the answer is already posted.

The quizzes at least should stand for a few hours of insolvency.

Secondly though I am posting as a guest i am a regular member. I am just a bit of the GA race (and looking at the quality of the GAs given) and number of posts race, that is the reason i am just not logging in.

The only credible GA I see is of Fyz (and I have a lot of respect for him and Milo).

And how does it matter

Thats why you hide your identity and prefered to remain as guest.?

had I logged in would you have known me from another ? is your avatar itself not an anonimity?

Grow up.

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Guru

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#22
In reply to #20

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 10:30 AM

I took a similar view about "not competing or becoming gurufied" for quite a time. Kris talked me out of it, effectively on the basis that at least people who recognised authors could make up their minds whether to ignore posts that were at first sight incomprehensible or to re-read them

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#23
In reply to #20

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 10:00 PM

when I take a challenge I don't look at the answers already made.

Good. But still you didn't post your answer, thats the fact. Thus only request is not to undervalue who have solved the problem.

had I logged in would you have known me from another ? is your avatar itself not an anonimity?

This is very true. With your Awatar and name, still you are unknown to all others unless you know the person personally and you know that so and so is Avatar of so and so person. Still one gets some identity in anonimity. Even though this is the fact, surprisingly people (including you and probably me too)prefer to hide their identity while making contravertial statements or while critisizing others. You did the same.

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Anonymous Poster
#24
In reply to #23

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/04/2009 5:07 AM

Why do they put quizzes answered within a few hours with no controversy?

Could you solve the problem? Fact is that you could not. Thats why you hide your identity and prefered to remain as guest. Then why underestimate those, who solved the problem.

making contravertial statements or while critisizing others

Who has carried out a unfounded criticism?

And the second (or the first part)

But still you didn't post your answer, thats the fact

Yes that is a fact. And (at least in my long bygone school/college days) I am not sure about the current schooling system, after the first student has already answered correctly we weren't encouraged to repeat the same. Did you like the number of "And the answer is 131" ?

Any grey spots, yes always open for discussion, but 2+2 ?

The only new insight, was of Fyz. There had been a few good ones earlier (Head light) , the answer of CR4 is controversial in that since that has been a very interesting subject as I remember, and pardon me, am not going to check back again. But had a fun in that since it went into the areas where different theories clash.

Let there be a proper challenging question that has a capacity to at least activate the brain cells. And i love the ones where multiple view points are possible.

In the end , with all due regards to Fyz, I am loving the guest avatar. And no, may be here there is a controversy, but as guest I have already accumulated more than 10 GAs in a month or so, and as I still see, it is much more difficult to get GA as a guest , since in a far more substandard reply (at least as I think) I got at least 25-30% of my hundred odd GAs where as here there was not even a passing mention, so that way Fyz is correct. In my avatar I would have been now somewhere around 125 or so.

But then are CR4 members ? A GA is a GA from wherever it comes. And without makin any mention, there are posts getting GAs that says refer to the previous posts (and the links attached) or the parts of the previous posts duplicated and the things on that aspect.

Fyz : As I see it the GAs are supposed to be the future reference points voted by the members - they should bring out something new, from members point of view (as the star rating of threads) - but things are sadly not on those lines. And that has a capacity of disillussioning me from the forum.

I am still trying to contribute, as a guest whenever something interesting comes of which I have some knowledge.

GSuhas: This is criticism, what I am doing now.

And since this aspect needs our fellow CR4ers to think over, I am putting on topic.

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Guru

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#25
In reply to #24

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/04/2009 12:49 PM

Well, Guest or not is a matter of taste. But I don't believe it makes much difference to GA votes. I actually get a much higher proportion when I don't declare anywhere who I am. I see this as totally pervers, given that in these cases the reason I don't declare is only that I'm uncertain I know what I'm talking about.

Note that: in the "daughters and sons" thread, only not-quite-answers and incorrect answers have been given GAs.

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Anonymous Poster
#35
In reply to #25

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/08/2009 5:13 AM

It does I didn't want to spell out. But the GAs are in this forum specific to the man answering.

One example?

OP: How can we differentiate between ordinary seamless pipe and IBR alloy steel pipe

#1 : Mill certificate. Any code ASME or IBR are going to need the original mill certificate. For non critical/ non major activities, you are free to use the other one, after ensuring the correctness that is.

#2:There are a lot of materials where the mill can manufacture the pipes, such as carbon steel, alloy steel, coper and copper alloys, nickel and nickel alloys, ... etc. The manufacturing methods are Seamless, Seam Welded and Spiral as shown.

Any one worth a GA ? (answering to the point? An answer is to a question and should have some new aspect brought in.).

I could have expanded - there are now devices to check the quality, which is done at our works to confirm randomly the supplies (despite the mill certificates) - the PMI devices and a lot of other paraphernalia. But since the OP is to code (IBR - Indian Boiler Regulations) then you do not need these - except like us to have a double check.(For IBR, the mill certificates are enough).

BTW: one of them was from me and though that did get a GA point, but i resent it. This is just common sense and must have been known to the OP too since he is manufacturing to code at least the question implies that.

And as manufacturer to code vessels (we are ASME U stamped, IBR, API monogram holder,...) We know, even our workers know about the PMI processes, and that is without exaggeration.

This is the aspect that has taken me out of the GA race. As a guest at least I am getting the extreme GAs (Good answers) tagged. Though I am missing a lot of GAs had i been logging in as my avatar, but then there are quite a few who had been voting me GAs by my name . And also some of the GAs as per me were not tagged but that is for the other members.

And GSuhas, when I find a GA, I will log in, mark it, and log out (Simple )

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#26
In reply to #24

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/04/2009 9:44 PM
Reply
Anonymous Poster
#27
In reply to #24

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/05/2009 10:36 AM

What is a GA? Not familier with the initials and what they stand for! What do you get for GA's?

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Anonymous Poster
#28
In reply to #27

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/05/2009 1:08 PM

Good Answers (You and I are not permitted to vote an answer as good) only registered members when logged in can mark an answer to be good (or the reverse is off topic)

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#29
In reply to #24

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/05/2009 9:53 PM

Hi my anonymous friend

You say, you got more than 125 GAs. We need to only believe you. As a guest these figure will never appear on anybodies screen.

As a guest you will get appreciations from others (GAs) and you will be proud of these. But as a guest you are not allowed to appreciate others, as you are not permitted to do so. It is shear injustice my friend. Please sign up and become regular member.

We all welcome you.

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#11

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 1:39 AM

There are 131 dealers. Mike sells 2403 apples to each dealer and has 34 left over. Juliet then has 1,199,567 which equals 9,157 per dealer. Bob would have 683,820 which equals 5,220 per dealer.

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#13

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 3:01 AM

Say m, j, b are the number of apples of Mike, Juliet and Bob respectively. Also let d be the number of dealers. As a programmer, I like to use the notation % for mod. So, the number of apples that Mike has to leave over, is m%d.

Now, it must hold (j + m%d)%d = 0 and (b + m%d)%d = 0 which are equivalent to:

(j + m)%d = 0 and (b + m)%d = 0 (see note * below for a proof)

Therefore, the sums j+m and b+m must both be multiples of d and consequently we need to see which of the sums' prime factors are common. These factors or any combination of their products will be candidates for d. Fortunately, in this case there is only one common factor, 131, i.e. the number of dealers is 131.

* Note: proof that (j + m%d)%d = (j + m)%d

We know that: m/d = [m/d] + (m%d)/d, so (j + m%d)%d = (j + m - [m/d]d)%d. The factor [m/d]d is a multiple of d, so it doesn't matter adding it or substracting it. Therefore, (j + m%d)%d = (j + m)%d

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#15

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 4:59 AM

I make it 131

Codey

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#16
In reply to #15

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 5:38 AM

Now they decide to have less dealers so that they have less contacts.

Thus they keep 5 apples for their family each and distribute all apples between 31 dealers.(or even have 127 or 131 dealers to maintain good relations with all existing dealers.

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Anonymous Poster
#19
In reply to #16

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 7:17 AM

According to my calculations, even if you distribute the total evenly you will have 29 apples left if there are 31 dealers.
You could have smaller residues with larger numbers of dealers, e.g. 3 with 43, 4 with 87, 21 with 125. However, I agree that 30 at 127 might be a suitable compromise.

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#18

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/03/2009 6:15 AM

just one , big government and the cash for dunkers program.

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Guru

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#30

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/06/2009 12:56 AM

i didn't understand the question itself:

is asked for

x(a+1)=314827

x(b+1)=1199522

x(c)=683786

?

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Anonymous Poster
#31

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/06/2009 6:03 AM

Product of M= of 314,827 apples, J = 1,199,533 apples, B=683,786 apples,

If No of dealers = D

Then the remainder of M/D (ie 314,827/D) say = Δ , if added to J , will make whole number if divided by D (and same with B)

ie mathematically

Rem [ ( J+ Rem (M/D))/D] = 0 and

Rem [ (B+Rem(M/D))/D] = 0

the way phys has solved is quite common sense (which is not common )

if rem(M/D) = Δ then

M = n1D+Δ where n1 is an integer

and since

Rem [(B+ Rem(M/D)/D] = 0 that means

D is a factor or (B+Δ)

ie it must be a factor of (B+Δ+n1D) too

ie it must be a factor of (B+M)

and the same logic for J too

ie D is a common factor of (J+M) too.

then it is simple - you find the HCF of (B+M) and (J+M) and that is the answer. To have the unique answer the D must be a prime number, else all the prime and non prime factors of the HCF could have been a valid answer.

(This is the mathematical explanation of the Phys cum the question)

Foot Note : Am I confused/ could I confuse you enough

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#32
In reply to #31

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/06/2009 10:31 AM

314827 + 1199533 + 683786 = 2198146

number of apples differ apd dif apd dif
dealers per to
dealer total
1 314827 0 1199533 0 683786 0
131 2404 -97 9156 97 5219 97

1 314827 0 1199533 0 683786 0
2 157414 -1 599766 1 341893 0

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Guru

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#33
In reply to #32

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/06/2009 10:34 AM

i hope the table is readable, there are some mising tabulators

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Anonymous Poster
#34

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/06/2009 6:41 PM

I'd like to know who or what counted 2 million+ apples and with what accuracy. If one of those numbers is just 1 apple off, it would completely change the answer.

This seems like an awfully easy question for being a monthly challenge - not that we will probably ever see the answer. None of the previous four monthly challenges have had answers posted.

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Anonymous Poster
#36
In reply to #34

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/08/2009 5:15 AM

Very interesting point (not the second one - no answers) the first one- who counted the two million plus.

What is the error in the count?

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Guru

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#37
In reply to #34

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/09/2009 1:05 AM

I think all 3 farmers have computer aided apple-counters, so that result should be exact (it's just an integer number).

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Anonymous Poster
#38

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/10/2009 5:05 PM

There are 131 apple dealers.

Mike and Juliet's production totals 1,514,360, and it must be an even multiple of the number of dealers, which would be an integer. Mike and Bob's production totals 998,613, and it also must be an even multiple of the number of dealers. The only integer factor shared by 1,514,360 and 998,613 is 131, as found in a calculator that came up in a google search for "Greatest Common Factor".

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Guru

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#39
In reply to #38

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/10/2009 5:12 PM

Given that Mike and Bob's production totals 998,613 (an odd number), presumably you are using "even" in a colloquial manner - that is to mean "integer".
(If this had been the first posting to introduce the method, I would have given it a G.A., regardless of syntax).

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Anonymous Poster
#44
In reply to #39

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/11/2009 9:22 AM

Yes, I did not mean even in the sense of multiple of 2. I meant even in the sense of integer.

Never posted here before. I saw the problem come thru on a spam e-mail, got interested, and solved it. Didn't know I wasn't or was the first to see it in this way, and didn't read the bazillions of prior posts before posting. Sorry if it offended you...

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Guru

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#45
In reply to #44

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/11/2009 10:11 AM

No offense taken (perhaps more important, none meant either).

Intelligent posters are always welcome. The only reason for my comment on "even" is that this sort of thing (and I've been guilty of worse myself) has a tendency to breed and then lead to massive confusion.

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Participant

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#41

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/10/2009 8:56 PM

Simple

The sum of Mike's and Julliet's apples (1514360) and also the sum of Mike's and Bob's apples (998613) can be divided exactly between the dealers.

I set up a quick excel worksheet with a dealers column numbering 1-200 and divided two totals by the number of theoretical dealers in two other columns. The lowest common denominator was 131 dealers.

Mike could sell 2403 apples each to 131 dealers (totals 31793) and could sell the remaining 34 apples to Juliet, who would then sell 9157 apples each to 131 dealers (totals 1199567). Or he could sell the 34 apples to Bob who could then sell 5220 apples each to 131 dealers (totalling 683820).

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Participant

Join Date: Nov 2009
Posts: 1
#43

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/11/2009 2:34 AM

Using a more democratic approach -If MIke had 4 extra apples over -Juliet and BOB could share Mikes extra apples between them 2 each -allowing all 3 farmers the opportunity to sellout all their stock to only three dealers

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Anonymous Poster
#46

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

11/19/2009 6:10 AM

139 dealers.

Mike has 131 apples left over,

Juliet can buy 37 and Bob can buy 94 apples.

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Anonymous Poster
#47

Re: Apple Dealers: CR4 Newsletter Challenge (November 2009)

12/01/2009 11:10 AM

I get 139

KL

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