Apple Dealers: CR4 Newsletter Challenge (November 2009)

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

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.

what about a small city with a population of 131?

and everyone is his own dealer?

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

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.

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

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

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

GA by any definition.

Thanks,

Jim

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...

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

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.

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.

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.

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

I make it 131

Codey

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.

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.

i didn't understand the question itself:

is asked for

x(a+1)=314827

x(b+1)=1199522

x(c)=683786

?

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}

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

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

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.

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

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".

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).

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...

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.

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).

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

139 dealers.

Mike has 131 apples left over,

Juliet can buy 37 and Bob can buy 94 apples.

I get 139

KL