Lil math puzzle

[IndiReloaded]

I suppose you could do some kind of area under the curve thing, but like you Wish, I didn't see the point.

Interested to see what DM says. I'm a long time out of math but I always loved the subject.

The formula I remembered btw, is what you figured intuitively (like Gauss did, lol):

a(n)=a1+(n-1)d

The d is the common difference between terms. Its the same form every student learns when starting to learn about sums & series.

How did you make the summation sign show up, DM?

[IndiReloaded]

Here's a good one from a math puzzle book. Show how you got your answer:
______________
The son of a rich bullion merchant left home on the death of his father. All he had with him was a gold chain that consisted of 123 links. He rented a place in the city center with a shop at the lower level and an apartment at the upper level. He was required to pay every week one link of the gold chain as rent for the place.

The landlady told him that she wanted one link of the gold chain at the end of one week, two gold links at the end of two weeks, three gold links at the end of three weeks and so on.

The son realized that he had to cut the links of the gold chain to pay the weekly rent. If the son wished to rent the place for 123 weeks, what would be the minimum number of links he would need to cut?

[GrkScorp]

anyone care for a beer?? i've got an extra 6 that i won't drink.

raverboy

An infinite number of mathematicians walk into a bar.. the first one talks to the bartender and says..

"Let me get a beer"

The bartender starts to get him a beer, but just as he does.. the second mathematician walks over and says..

"Let me get half a beer"

The bartender reaches for a second beer, but just as he does.. the thrid mathematician walk over and says..

"Let me get 1/4th a beer"

The bartender looks at the endless number of mathematicians filling up the bar.. he puts two beers on the bar counter and says..

"You're all @ssholes, enjoy"

[DoesntMatter]

LOL, this^ is what I said DM.

How? you said
S = 1 + 2 + .....

S is not number of blocks, it is the step number. Unless you are using S where I used N, but in even such case evaluating the sum as such is cumbersome

lol. You don't need calculus or any formula to solve this question. I used a rather primative method. I just visualized the steps as a full cube and saw that there are patterns as you increase the number of steps. This is from a humanity and social science student. Not a math and life science person.

No you don't need calculus but the fact that it can be (kind of) used makes it kind of interesting. The reason I liked this problem is because there were quite a few ways to solve it, and it's interesting to see how different people solve it

But you solved this by imaging the stairs as a CUBE? I can see how you would do it as a SQUARE... If you draw in squares to make the stair figure a square, you can see that

N(s) = s + N(s-1)

and

N(s) = s^2 - N(s-1)

So solving the system of equation gives the same thing

But what do you mean you saw this as a cube?? Like the figure was a quarter of a pyramid or something?

I suppose you could do some kind of area under the curve thing, but like you Wish, I didn't see the point.

Interested to see what DM says. I'm a long time out of math but I always loved the subject.

It doesn't really *use* calculus perse but draws upon some of it's concepts. It's a pain in the ass and not very intuitive method



OK though now Indi's problem-

Clearly I'm missing something if the guy pays 1 additional link per week from a chain with 123 links and desires to stay 123 weeks since he doesn't have enough.

But once you clear this up is the chain linear or circular like a necklace someone would wear? And are "cut" links payable? And if a link is cut and one side payed, is the "cut" linked considered removed from the second half of the chain?

[IndiReloaded]

How? you said
S = 1 + 2 + .....

S is not number of blocks, it is the step number. Unless you are using S where I used N, but in even such case evaluating the sum as such is cumbersome

I only gave it as an example of an arithmetic series, DM, I didn't define anything. I was trying to understand the basic form of the problem. The sum that was solved for IS of the form 1 + 2 + 3 ... as I said, its the most basic type of arithmetic series.

Anyway, good question & its solved now. I think I'll give it to my son & see if he's as smart as Gauss was.

Still want to know how you got the sigma symbol to appear in your post. My keyboard doesn't have that symbol.

My problem isn't complicated. He has 123 links and 123 weeks. One link per week to the landlady, just like the problem says.

What's the min number of cuts to the chain to pay in full? Linear chain, cut links payable.

[DoesntMatter]

Do you mean this? --> Σ

Well then to answer your problem I would say he doesn't have enough links

[IndiReloaded]

[quote=DoesntMatter;322927]Do you mean this? --> Σ

Arrrgh!!! How you DOOOOO that??

Well then to answer your problem I would say he doesn't have enough links

He has enough links. 123 links, 123 weeks. Try again, don't think too hard. The middle paragraph confused me at first, too, but if you read it exactly as is, it makes sense.

[DoesntMatter]

For the sigma character I just copy and pasted it from [url]http://en.wikipedia.org/wiki/Sigma[/url]

Is this some type of trick question? Because on week 123 he has to pay 123 links, and on week 1 he has already payed 1 of his 123 so now he has 122.

OK Indi I told you where the sigma character comes from now you tell me this!

[IndiReloaded]

DM, you are thinking about this too hard. No, its not a trick question.

The first week, he pays the lady 1 link. She has 1 link at end of wk1.

The second week, he pays the lady 1 link. She now has 2 links at end of wk2.

The third week, he pays the lady 1 link. She now has 3 links at end of wk3.

And so on until he has paid all 123 links for 123 weeks.

What is the fewest number of cuts he needs to make?

HINT: If you cut a link in the center of the chain, you get three lengths: a one-link piece and two other lengths.

Get it?

[Illusional]

well 6 - 6 = 0. which is the total number of beers that i have left from last night.

raverboy

[DoesntMatter]

Well that makes it a lot easier Indi now that I know he isn't paying an increasing amount of links per week. I thought he had to pay the bitch an extra link every week, in which case he would not have enough links

If he cuts the link in the center every time, the length of the 2 new chains are half of 1 less than the old chain length. If you consider each new collection of chains of same length (or nearly the same length, 1 off if parent length is even) as belonging to a generation, say generation n, the length L of the chain is determined by the generation by

L(n) = [L(n – 1) – 1] / 2

The final length we are looking for is L = 1, so backsolving this shows you need n = 6, or 6 generations of lengths

Also, you notice we start with 1 chain and the number of cuts required to get to the next generation is 1. Getting to the next generation requires 2 cuts, and the next 4, since each cut produces 2 new chains. So the number of cuts increases exponentially according to 2^n, with n being the generation. The first generation when number of chains = 1 requires only 1 cut, so this must be generation n = 0. Six generations are needed, so sum up the number of cuts from n = 0 to n = 5, which is 2^0 + 2^1 + … + 2^5 which gives you 63 cuts needed

What method did you use Indi?

[DoesntMatter]

Haha riddle me jokester says this observer!

I'll do this tomorrow when I'm no longer buzzed and it's not 2am, last time I read your puzzle I hadn't eaten for around 30 hours and we all know what happened then

[DoesntMatter]

WTF!?? You just had a post here, where did it go!

[IndiReloaded]

Yes, that is the easier version of the problem b/c I was having trouble explaining to you. Okay, now that's sorted out, here's the rest:

Same problem. Guy has to pay her 123 links over 123 weeks.

NOW he doesn't necessarily pay her a *single* link a week, only that he pays her the correct *increment*. I.e. at the end of each week she should have:

w1 = 1 link given
w2 = 2 links given
w3 = 3 links given
w10 = 10 links given...

and so on. Now, what is the minimum # of cuts (its a lot less than 60).

Hint: you can take back/exchange links.

This is the original problem. Hope this makes more sense.

[IndiReloaded]

WTF!?? You just had a post here, where did it go!

I made a mistake & amhaving trouble w/my wireless, DM. May need to come back later. Here it is again, babe.