Cool, another thread inspired by the Sea.
I used to like Math.
Until I fucking dropped it to concentrate on manga.
The Mathematics Thread
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I've got a pretty sweet math problem that involves more thinking and less calculating.
Suppose you have a bag of gems. When you take the gems out of the bag two at a time then there is one left over at the end. Similarly if the gems are taken out three, four, five, or six at a time, there are two, three, four, and five gems left over respectively. Annoyed, finally you take the gems out of the bag seven at a time, and no gems are left over. How many gems are in the bag?
Edit: Lol. I just realized that if this was Professor Layton the answer would be zero since I sort of asked how many gems were in the bag after you took them out. I should have said "how many gems were in the bag"?
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I've got a pretty sweet math problem that involves more thinking and less calculating.
Suppose you have a bag of gems. When you take the gems out of the bag two at a time then there is one left over at the end. Similarly if the gems are taken out three, four, five, or six at a time, there are two, three, four, and five gems left over respectively. Annoyed, finally you take the gems out of the bag seven at a time, and no gems are left over. How many gems are in the bag?
Edit: Lol. I just realized that if this was Professor Layton the answer would be zero since I sort of asked how many gems were in the bag after you took them out. I should have said "how many gems were in the bag"?
Hmmmmmmmmmmmmmm
mmmmm
mmm
is the answer 119?
119 is dividable by 7
118 is dividable by 2 (1 left over)
117 is dividable by 3 (2 left over)
116 is dividable by 4 (3 left over)
115 is dividable by 5 (4 left over)
114 is dividable by 6 (5 left over) -
77 almost works.
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almost… but not quite
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Hmmmmmmmmmmmmmm
mmmmm
mmm
is the answer 119?
119 is dividable by 7
118 is dividable by 2 (1 left over)
117 is dividable by 3 (2 left over)
116 is dividable by 4 (3 left over)
115 is dividable by 5 (4 left over)
114 is dividable by 6 (5 left over)Indeed you are correct!
A quick way to see the answer is to realize that if you add one gem to the bag, then the number of gems is divisible by 2,3,4,5 and 6. So the number of gems plus one is a multiple of 60 (223*5). Combining this with the fact that the number of gems is divisible by 7 will quickly give you 119 as the smallest possible answer.
New problem:
Suppose you are at a market and you want to buy a specific number of pumpkins. There are just three merchants at the market who sell pumpkins and due to the recent pumpkin harvest they will only sell pumpkins in lots of a specific number. The first merchant will only sell pumpkins in groups of six, the second merchant will only sell pumpkins in groups of nine and the last merchant will only sell pumpkins in groups of 20. Using these three merchants, is there a largest number of pumpkins that it is impossible to leave the market with? If so, what is this number? -
Indeed you are correct!
A quick way to see the answer is to realize that if you add one gem to the bag, then the number of gems is divisible by 2,3,4,5 and 6. So the number of gems plus one is a multiple of 60 (223*5). Combining this with the fact that the number of gems is divisible by 7 will quickly give you 119 as the smallest possible answer.
Wrote a little list comprehension for finding solutions under a given value in Haskell. I tried it with 10000 as a limit. Seems to work.
let gems y = [x | x <- [1..y], x
mod
7 == 0, (x+1)mod
60 == 0]
gems 10000
[119,539,959,1379,1799,2219,2639,3059,3479,3899,4319,4739,5159,5579,5999,6419,6839,7259,7679,8099,8519,8939,9359,9779] -
Nice. I have a program that solves the pumpkins problem too if that inspires anyone.
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… understands nothing.
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Anyone using Matlab?
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drE5cHe6c3s
I couldn't find a more fitting thread for this.
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I was thinking more Donald Duck in Mathemagical land XD
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Anyone using Matlab?
I think i've started it a few years ago. Though i won't be able to help you if you have a specific problem. I have worked a bit with maple thus i should know its basics.
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drE5cHe6c3s
I couldn't find a more fitting thread for this.
I love Look Around You so much. At least the first season.
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drE5cHe6c3s
I couldn't find a more fitting thread for this.
45000000001?
XD
That was great -
It's the future and Queen Elizabeth the III and Queen Elizabeth the IV are going to a party hosted by Queen Elizabeth the V. They're keen to make the right impression so it's important they choose their outfits carefully.
Queen Elizabeth III has 40 dresses to choose from, whereas Queen Elizabeth the IV has 4000.
Queen Elizabeth the V has just one dress.
But it has the ability to transform itself into the shape of any dress.
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That is simply the best math problem I've heard in my life.
"The party was canceled."
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@Holy:
It's the future and Queen Elizabeth the III and Queen Elizabeth the IV are going to a party hosted by Queen Elizabeth the V. They're keen to make the right impression so it's important they choose their outfits carefully.
Queen Elizabeth III has 40 dresses to choose from, whereas Queen Elizabeth the IV has 4000.
Queen Elizabeth the V has just one dress.
But it has the ability to transform itself into the shape of any dress.
Elizabeth 3 & 4 can just choose the dresses they like… Elizabeth 5 doesn't require a dress, she will go to the party nude & pose for artists with 3 & 4. Thus stimulating the new begining of 'The Renaissance'...
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BlvKWEvKSi8
This guy knows what he's talking about!