Discussion in 'Physics & Math' started by ProCop, Jan 8, 2005.
S, M, T, W, T, F, S
J, F, M, A, M, J, J
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Ah well, I didn't think it would take long. Please Register or Log in to view the hidden image!
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is it 65511?
i apologize if this and the last puzzle have already been solved i didnt notice there are so many pages i didnt look for the answer in the other posts this is what i came up with 1,1,3,5,8,13,20,34,75,145,298 ill explain how but i really cant do it through an equasion, never got that far in school so if im right and you want me to explain,itll look really infintile. :m: Please Register or Log in to view the hidden image!
i was just looking through the other pages, saw that some one already solved it i actually was surprised i got it right. if you have time could you tell me if the equasion that was used to solve the puzzle came to that number in essence the way i did here is how i came up with it
the sequence seemed to revolve aroundthe previous numer multiplied by 2 thena number either being added or subtracted to the sum to come up with the next number. the number added or subtracted was a multiple of 2 and they seemed to be combined to make 4's so if you take the final number in the sequence 32747, it is the previous number(16367multiplied by 2+13) and the ones prior went (previous multiplied by 2+13), p2+15, p2+11. see 4's. so with the final #32747multipliedby 2+13+4=65511? Please Register or Log in to view the hidden image!
Take the list of words below and arrange them into 3 sentences that all have something in common. Each word is only used once each time they appear in the list. ('A' can be used 4 times as it's in the list 4 times.) Punctuation is not an issue.
A, A, A, A, ANIMALS, BAR, BAT, DAN, DROOP, I, I, IN, IN, IS, IT, NET, OR, POOR, SAW, SLAM, TEN, WAS
What are the sentences and what do they have in common?
I haven't worked out the sentences yet, but they're all palindromic.
TEN ANIMALS I SLAM IN A NET
POOR DAN IS IN A DROOP
WAS IT A BAR OR A BAT I SAW
Sarkus is right. Nice job.
(and geodesic answered the question of what they had in common, although I'm not entirely sure how he answered it before knowing what the sentences were).
i m sofa king we todd id
I just noticed the pairs of words which were obvious - animals/slam in a, droop/poor, and made an educated guess. Please Register or Log in to view the hidden image!
Not mine, but it's interesting:
“To reward you for killing the dragon,” the Queen said to Sir George,
“I grant you the land you can walk around in a day.” She pointed to a pile
of wooden stakes. “Take some of these stakes with you,” she continued.
“Pound them into the ground along your way, and be back at your starting
point in 24 hours. All the land in the convex hull of your stakes will be
yours.” (The Queen had read a little mathematics.)
Assume that it takes Sir George 1 minute to pound a stake and that he
walks at a constant speed between stakes. How many stakes should he take
with him to get as much land as possible?
A simpler one:
From England comes the series : : : 35, 45, 60, x, 120, 180, 280, 450, 744,
1260, : : : Find a simple continuous function to generate the series and compute
the surprise answer for x.
- Both from 'Mathematics Puzzles from Around the World'
I should note: Any riddle I post on here is not my own and will most likely be found on the internet. I don't mean to take credit for any riddle I post, and if it is my own, I will say so.
That's all I guess.
Ah - a simple mathematical formula is required... Please Register or Log in to view the hidden image!
Okay, the obvious answer is that the largest area will be a polygon with N number of stakes. Each stake takes 1 minute to pound, and there are 1440 minutes in total - so walking time in total = (1440-N) minutes.
Okay - the length of each side of the polygon will be vel*(1440-N)/N - being the total walking time * speed divided by the number of sides.
The angle going from one peg to the centre to the next peg is 360/N (in degrees, or 2*pi/N in radians).
The area of the triangle described by two adjacent pegs and the centre is thus: 1/2 * base * Vertical height
We know base is vel * (1440-N)/N
We also know that the vertical height can be known as Tan X = 1/2 base / vertical height where X is half of 2*pi/N = pi/N.
So Vertical Height = 1/2 base * COT X
= vel * (1440-N)/2N * COT (pi/N)
We get that the area, for a given N, is defined by:
N * 1/2 * vel*(1440-N)/N * vel*(1440-N)/2N * COT(pi/N)
= vel^2 * (1440-N)^2 * COT(pi/N) / 4N
Now, we want the N that maximises this.
The easiest way is with a spreadsheet.
So doing this from N=1 and above we get the biggest area being when N=17 and the area is 159,300.2*vel^2.
A little quasi-emprical investigation on an Excel spreashdeet shows that n=17 is indeed the theoretically correct answer. However, Sir George's land-grab will be affected by barely 1% if he chooses as few as 8 or as many as 30 stakes.
On the other hand, if he spends even a short time eating, sleeping, urinating, or defecating instead of walking, his total area will drop dramatically. I'm going to be a nuisance and suggest that the improvement in Sir George's walking efficiency by carrying a few less stakes will outweigh the losses involved.
On the whole, I think that n=8 is the best number, not the least because of the simplicity in navigation. Consider how much harder it would be for Sir George to accurately plot the corners of a seventeen sided polygon than an eight sided one!
What was this doing on the second page?
Anyway, a new riddle:
A lady had her birthday last week. At the party, she said "I have three children. The sum of their ages is the same as my age, and the product of their ages is equal to the year in which I was born."
How old are her children?
hmm... let me organize my thoughts...
Solve for positive integers a,b,c,n:
a+b+c = n
a*b*c = 2004 - n
25 < n < 105
15 < n - a < 50
15 < n - b < 50
15 < n - c < 50
Isn't it 2005 this year? Shouldn't a*b*c = 2005 - n ??
Or are we assuming this puzzle was set in 2004?
Okay - if it is 2004 - then the answer is: ages: 25, 26 and 3.
The sum is 54.
The product is 1950.
Year of birth is 2004 - 54 = 1950
The puzzle was unfortunately set in 2005, so that's not the answer I have.
Separate names with a comma.