#1: Sylvain "WCPman" (qwerty) on Oct 12, 2008

Who will risk a first anser to enigma #1????

Unless the part about "all the girls agree that June is in her prime" is supposed to be some kind of clue, I think there are lots of possible solutions to this, even assuming that each letter was assigned a non-negative integer. For one thing, each name has one J and one E, so the values of J and E can always be interchanged.

One possible solution is N=1, A=2, U=3, Y=4, E=5, and J=8.

Another is N=3, A=1, U=2, Y=0, J=5, and E=7. I think there are zillions.

I'm thinking that "June is in her prime" is a clue that perhaps all of June's numbers are prime numbers...? Haven't sat down to play with it yet... my head is pounding...

I think J and E are 2 and 3, n=1, u=11, y=12 and a=10

I'm at work right now but I'll give you a small hint later tonight

But "June is in her prime" is a hint I'll give you that

That interpretation of "June is in her prime" can't give a unique answer. I already pointed out that the values for J and E can always be interchanged, and they are both in "June".

hint hint hint

the "june is in her prime" tell us that all letter of the June name are prime number

plus all number are under 10

with that you should be able to answer the question

I'm assuming you also want all letters to have different values.

Here's one:

J=3, U=7, N=5, E=2, A=6, Y=4

Here's another:

J=2, U=7, N=5, E=3, A=6, Y=4

I don't know if there are other solutions. There are lots if you don't require all number to be different. I don't have a very systematic solution technique.

darn, if all numbers are under 10, then my first guess was wrong

ok, I'm with Jan's #8 on this one

Congratulation Jan the answer I got is

J=2 U=7 N=5 E=3 A=6 and Y =4

you got it right and in less than 2 days

Next one going to be harder for you I hope....

Impossible to say, since you never gave any units for measurements. Also, I don't know if the fact that there is a road in between the areas of land has anything do with it (like if he were to include the road with one of the sections). Also, I don't know if the fence needs to be higher on one of the sections of land or not. Maybe the cows won't jump over a 2 foot high fence. Maybe he wants a 20' high fence to keep kids or other animals out of the garden.

So my answer for how much fencing he will need is:

245 and 3/8.

Don't ask me what units, since you never told ME! :-)

the height is not a factor, consider it one section hight as for the unit its just a situation whatever unit you put in will give you the same answer

the perimeter can be in meter or in foot it doesnt matter since we are not looking for equivalent of one visit each other. Take the fact you have in pure mathyematic unit and put the unit you want at the end it was just a situation

the road element is to show that both area are split from each other so basically you have two rectangle of diffrent shape.

Let the first field be A by B, and the second field be C by D.

The first is eight times larger than the second.
A*B = 8*C*D

The second needs twice as much fencing to surround as the first.
2*(A+B) = C + D

We want to know the total amount of fencing needed
T = 2*(A+B+C+D)

-------------------

Just looking at it, I agree with Adam: T can be anything because once you find one solution then you can multiply the numbers by any constant to have another solution.

But from the response above, it sounds like Sylvain actually wants to know the proportions of the two fields, not their total perimeter.

We can eliminate the scale problem by just declaring that A=1. Then you can solve the two equations to get

C = (D - 2) / (16*D - 1)
B = 8 * C * D
A = 1

We'll clearly need to pick D as a number greater than 2 so that C will be greater than zero, but any number greater than two will suffice. So even eliminating scaling, we have an infinite number of solutions.

Here's two. I generated them from D=3 and D=4 in the above equations and scaling the whole thing to eliminate fractions:

big field: 24 x 47 ; area= 1128 ; perimeter = 142
small field: 1 x 141 ; area= 141 ; perimeter = 284

big field: 63 x 64 ; area = 4032 ; perimeter = 254
small field: 2 x 252 ; area = 504; perimeter = 508

I can generate lots of other solutions.

Jan I'm impress by how fast you solve this one too I'll have to try something harder for next week

your second guess ( 254 and 508) is the good answer if you add both of them togeter to have the total number of fencing to buy that is

great job

Why is that one good and the other one not?

Both meet the requirements of the puzzle, and so do an infinite number of other combinations of numbers, which all add up to different amounts of fencing.

that was simply the answer I got but your right both are good answer

+*/+ =10

or

++-+ =10

you can use each only once
and that not the good answer

It is 2+3x4/5+6 = 10
2+3 = 5, then 5x4 = 20, then 20/5 = 4, then 4+6 = 10
Although mathematically, the order of operations would require that I multiply and divide before I add or subtract.
But to follow your rules and do it straight from left to right, this is the only way.

Finally I got one first! :-)

no you don't

you have to use all 4 sign once

try again

Are we following the order of operations? Cause if we do that, Adam's answer ends up being 11.04 3*4 = 12, 12 / 5 = 2.04, 2.04 + 3 = 5.04, and 5.04 + 6 = 11.04.

Oops. Never mind. LOL

Assuming we do the operations from left to right
2/3-4+5*6
That's actually the first thing I tried, meaning to eliminate the possibility that the / goes first, but that could still give an integer if we multiplied by the 6, so I decided to check that out, and lo and behold, it worked.

I'm pretty sure that that is the only possible solution. There are only 24 possibilities after all.

If you are doing multiplication and division before addition and subtraction, then I'm pretty sure that no combination gives you ten, or even an integer.

Actually, that comes to 10.000002. Is that really "10?"
LOL
I also had "eliminated" that the "/" went first, since that gave a fracture, but I never thought to check that by multiplying it by 6 you could get a whole integer again. Jan is the man!

man you're a math machine Jan you have it again

great job

H?

YES

that was fast

great job

Wow :D I finally got something right :) thank you

I started backwards and I think the guy had $1960

ggreat job you got it

I guess you'd write the formula like this:

((((x-100)/2-10-100)/2-10-100)/2-10-100)/2-10 = 10

Here, x is the amount of money he started with. Each night he first spends 100 to get in, then loses half of what he had left, and then spends 10 going out. We just need to solve this for x:

((((x-100)/2-110)/2-110)/2-110)/2 = 20
(((x-100)/2-110)/2-110)/2-110 = 40
(((x-100)/2-110)/2-110)/2 = 150
((x-100)/2-110)/2-110 = 300
((x-100)/2-110)/2 = 410
(x-100)/2-110 = 820
(x-100)/2 = 930
x-100 = 1860
x = 1960

And, yup, I get the same answer. Sounds like Merili skipped the algebra and just imagined time going backward. The guy walks backwards into the casino with 10 dollars in his pocket, the coatcheck girl gives him 10 more dollars bringing him to 20, he doubles his money to 40 dollars, then gets paid 100 to walk backwards out of the casino, and so on. Maybe visualizing it that way is more fun than just doing the algebra, though it is the same thing really.

lol Jan :D That was pretty much how I imagined it :D

I did the exact same thing as Merili, and also got $1960. You just beat me to it!

I was going the same way as Merli and Adam, but I was counting the $100 entrance fee as part of the money he lost.

I didn't even think of that. Very good point. In that case, couldn't you also count the $10 as money he lost too?

Yay, I beat Adam :D Didn't think that was possible :)

Oh come on, don't say that. This was the only one of these I have actually gotten right. So why would you think that?

Actually by beating you I meant being faster than you :)

Yeah I was counting the 10 as money he lost too.

I think it's 15 chickens and 10 ducks. I didn't really remember how to solve these kinds of problems, so I just drew a graph with the two constraint lines (barnyard space and poultry price). This gave me a four-sided polygon with four vertices (C=0,D=0), (C=0,D=25), (C=15,D=10), (C=20,D=0). I knew that the maximum would have to be at a vertex (since we are lucky enough that they are all integers), so I evaluated the eggs per week function at each vertex, getting 0, 100, 110, and 100 respectively, so the third vertex was my maximum.

once again you got it right and you got it first

I started with 20 chickens (maximum #, with only $80) which gave 100 eggs per week. If he bought the maximum # of ducks, which is 25, he also get 100 eggs per week, but he also saves $30 doing that. So I kept adding and subtracting chickens/ducks until I came up with the answer Jan had as the best solution, although he actually gets 115 eggs per week.

With 15 chickens ($4 each, = $60), at 5 eggs per week, he gets 75 eggs from them each week. Add that to the 40 eggs he gets from 10 ducks ($2 each = $20), and you get 115 eggs per week, not 110. Isn't that right?

Just a side note... if he can get 100 eggs per week from either 20 chickens (all of his $80) or 25 ducks (only $50 of his $80) would it be worth it to him to save the $30 for something else? Does it cost more to feed 25 ducks vs 20 chickens? If so, then maybe that $30 saved goes to feeding the 5 more ducks. Any thoughts?

this is only a math problem not a real most saves poor farmer situation Adam.....

lol

Huh?

Sorry, I didn't get what you just said.

it just that you consider real situation like

"more profitable if he don't spend all money" and stuff like that in a pure math problem included in a small story

Response #45 said: "This is only a math problem, not a real must-save-poor-farmer situation, Adam." Figuring out Sylvain's messages can be a little puzzler of it's own (though not nearly as difficult as figuring out my French would be).

I don't actually agree. Once we've solved the puzzle, it's time to look for more puzzles outside the puzzle. But I can't really get into the problems of this farmer, who seems to have a rather tiny farm and who has forgotten to account for the cost of feed and who seems to be housing chickens and ducks interchangably in the same space. I think he's hopeless, and should just sell his farm to condo developers.

and make a lot more money that way

rigth again Jan rigth again lol

But my original question was, it really is 115 eggs, right?

its 15 and 10 for the number of chicken and ducks

Yes, I misremembered the number of eggs when I typed it in my solution method. It is 115.

Look like this time its harder no first day answer ....

good thing

Still no answer ????

it that a hard one or people show they lack of interest suddenly ???

I wont post new brainbuster if no one wants them tell me

Somehow I never run across this when I have time on my hands.

It is not that difficult.
C=Cars, T=Trucks
Basic Linear Formula (15C)+(30T)=6000
Solved for C to find the number of cars with know number of trucks: C=-2T+400

1. Most amount of spots
Spots for cars: 300
Spots for trucks: 50
Total: 350 spots
Just set the number of trucks to the minimum of 50

2. Maximum revenue
Spots for cars: 200*$2=$400
Spots for trucks: 100*$5=$500
Maximum Revunue: $900
This involves having the most amount of trucks possible and still having twice as many cars, or C=2T
Our two equations are;
C=2T
C=-2T+400, combined

2T=-2T+400
4T=400
T=100

Goto next topic

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