[math-fun] Simple problem
A box holds 37 cubic inches. Add an inch to each side and it holds 83 cubic inches. Add another inch to each side and it holds 157 cubic inches. Add yet another inch to each side, and how much does it hold?
My method (use fixed-width font for viewing): 37 83 157 ? 46 74 28 6 Has to be a 6 down there, to make the leading coefficient 1. Thus: 37 83 157 265 46 74 108 28 34 6 Answer: 265. --Joshua
Hello, this is the finite differences of Newton, n^3 = (1,7,12,6) n^2 = (1,3,2) n = (1,1) 1 = (1) then if we have (37,46,28,6) as coefficients of the differences, a linear combination computation gives ; n^3 + 8*n^2 + 15*n + 13 37, 83, 157, 265, 413, 607, 853, 1157, 1525, 1963, 2477, 3073, ... the coefficients of n^k are related to Stirling numbers. ref : The book of Jordan < calculus of finite differences>. http://books.google.com/books?id=3RfZOsDAyQsC&printsec=frontcover&dq=finite+... best regards, Simon Plouffe Le 08/avr./2011 05:09, Joshua Zucker a écrit :
My method (use fixed-width font for viewing):
37 83 157 ? 46 74 28 6
Has to be a 6 down there, to make the leading coefficient 1.
Thus:
37 83 157 265 46 74 108 28 34 6
Answer: 265.
--Joshua _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Seems you could use the first two numbers to determine you have essentially a cube: (37^(1/3)+1)^3 = 81.3, which is pretty close to 83. Next two boxes are predicted 152 and 254, actual 157 and 265. Interesting that the error climbs to 5% then falls as the box gets more and more cube-like. 2% 81, 83 4% 152, 157 4% 254, 265 5% 394, 413 5% 578, 607 5% 813, 853 5% 1103, 1157 5% 1455, 1525 5% 1876, 1963 5% 2370, 2477 4% 2944, 3073 On Thu, Apr 7, 2011 at 8:29 PM, Simon Plouffe <simon.plouffe@gmail.com>wrote:
Hello,
this is the finite differences of Newton,
n^3 = (1,7,12,6) n^2 = (1,3,2) n = (1,1) 1 = (1)
then if we have (37,46,28,6) as coefficients of the differences, a linear combination computation gives ; n^3 + 8*n^2 + 15*n + 13
37, 83, 157, 265, 413, 607, 853, 1157, 1525, 1963, 2477, 3073, ...
the coefficients of n^k are related to Stirling numbers.
ref : The book of Jordan < calculus of finite differences>.
http://books.google.com/books?id=3RfZOsDAyQsC&printsec=frontcover&dq=finite+...
best regards,
Simon Plouffe
Le 08/avr./2011 05:09, Joshua Zucker a écrit :
My method (use fixed-width font for viewing):
37 83 157 ? 46 74 28 6
Has to be a 6 down there, to make the leading coefficient 1.
Thus:
37 83 157 265 46 74 108 28 34 6
Answer: 265.
--Joshua _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Ah - probably not your point, David, but this can't be a real box. No box would ever have volumes 37, 83, 156 as you increase the length of each side by 1. (At least one that's not non-imaginary.) You can get close to your series — volumes 36, 83, 156, 263 — starting with a box 4.25 by 4.25 by 2, for example. On Fri, Apr 8, 2011 at 12:10 AM, Gary Antonick <gantonick@post.harvard.edu>wrote:
Seems you could use the first two numbers to determine you have essentially a cube: (37^(1/3)+1)^3 = 81.3, which is pretty close to 83. Next two boxes are predicted 152 and 254, actual 157 and 265.
Interesting that the error climbs to 5% then falls as the box gets more and more cube-like.
2% 81, 83 4% 152, 157 4% 254, 265 5% 394, 413 5% 578, 607 5% 813, 853 5% 1103, 1157 5% 1455, 1525 5% 1876, 1963 5% 2370, 2477 4% 2944, 3073
On Thu, Apr 7, 2011 at 8:29 PM, Simon Plouffe <simon.plouffe@gmail.com>wrote:
Hello,
this is the finite differences of Newton,
n^3 = (1,7,12,6) n^2 = (1,3,2) n = (1,1) 1 = (1)
then if we have (37,46,28,6) as coefficients of the differences, a linear combination computation gives ; n^3 + 8*n^2 + 15*n + 13
37, 83, 157, 265, 413, 607, 853, 1157, 1525, 1963, 2477, 3073, ...
the coefficients of n^k are related to Stirling numbers.
ref : The book of Jordan < calculus of finite differences>.
http://books.google.com/books?id=3RfZOsDAyQsC&printsec=frontcover&dq=finite+...
best regards,
Simon Plouffe
Le 08/avr./2011 05:09, Joshua Zucker a écrit :
My method (use fixed-width font for viewing):
37 83 157 ? 46 74 28 6
Has to be a 6 down there, to make the leading coefficient 1.
Thus:
37 83 157 265 46 74 108 28 34 6
Answer: 265.
--Joshua _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
So 42 cubic inches is a correct answer to the original question, correct? Or at least as valid as the other answers? David, would you mark "42 cubic inches" as correct? This reminds me of the little-known fact that if you hold up a guinea pig by its tail, its eyes fall out (although strangely pet stores seldom warn of this). -tom On Fri, Apr 8, 2011 at 4:12 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Ah - probably not your point, David, but this can't be a real box. No box would ever have volumes 37, 83, 156 as you increase the length of each side by 1. (At least one that's not non-imaginary.)
You can get close to your series — volumes 36, 83, 156, 263 — starting with a box 4.25 by 4.25 by 2, for example.
On Fri, Apr 8, 2011 at 12:10 AM, Gary Antonick <gantonick@post.harvard.edu>wrote:
Seems you could use the first two numbers to determine you have essentially a cube: (37^(1/3)+1)^3 = 81.3, which is pretty close to 83. Next two boxes are predicted 152 and 254, actual 157 and 265.
Interesting that the error climbs to 5% then falls as the box gets more and more cube-like.
2% 81, 83 4% 152, 157 4% 254, 265 5% 394, 413 5% 578, 607 5% 813, 853 5% 1103, 1157 5% 1455, 1525 5% 1876, 1963 5% 2370, 2477 4% 2944, 3073
On Thu, Apr 7, 2011 at 8:29 PM, Simon Plouffe <simon.plouffe@gmail.com>wrote:
Hello,
this is the finite differences of Newton,
n^3 = (1,7,12,6) n^2 = (1,3,2) n = (1,1) 1 = (1)
then if we have (37,46,28,6) as coefficients of the differences, a linear combination computation gives ; n^3 + 8*n^2 + 15*n + 13
37, 83, 157, 265, 413, 607, 853, 1157, 1525, 1963, 2477, 3073, ...
the coefficients of n^k are related to Stirling numbers.
ref : The book of Jordan < calculus of finite differences>.
http://books.google.com/books?id=3RfZOsDAyQsC&printsec=frontcover&dq=finite+...
best regards,
Simon Plouffe
Le 08/avr./2011 05:09, Joshua Zucker a écrit :
My method (use fixed-width font for viewing):
37 83 157 ? 46 74 28 6
Has to be a 6 down there, to make the leading coefficient 1.
Thus:
37 83 157 265 46 74 108 28 34 6
Answer: 265.
--Joshua _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Check out Golly at http://golly.sf.net/
Tom — I think the original problem is interesting by itself. I just mean that if you graph the following there's no place where all three surfaces intersect. xyz=37 (x+1)(y+1)(z+1)=83 (x+2)(y+2)(z+2)=157 On Fri, Apr 8, 2011 at 4:30 PM, Tom Rokicki <rokicki@gmail.com> wrote:
So 42 cubic inches is a correct answer to the original question, correct? Or at least as valid as the other answers?
David, would you mark "42 cubic inches" as correct?
This reminds me of the little-known fact that if you hold up a guinea pig by its tail, its eyes fall out (although strangely pet stores seldom warn of this).
-tom
On Fri, Apr 8, 2011 at 4:12 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Ah - probably not your point, David, but this can't be a real box. No box would ever have volumes 37, 83, 156 as you increase the length of each side by 1. (At least one that's not non-imaginary.)
You can get close to your series — volumes 36, 83, 156, 263 — starting with a box 4.25 by 4.25 by 2, for example.
On Fri, Apr 8, 2011 at 12:10 AM, Gary Antonick <gantonick@post.harvard.edu>wrote:
Seems you could use the first two numbers to determine you have essentially a cube: (37^(1/3)+1)^3 = 81.3, which is pretty close to 83. Next two boxes are predicted 152 and 254, actual 157 and 265.
Interesting that the error climbs to 5% then falls as the box gets more and more cube-like.
2% 81, 83 4% 152, 157 4% 254, 265 5% 394, 413 5% 578, 607 5% 813, 853 5% 1103, 1157 5% 1455, 1525 5% 1876, 1963 5% 2370, 2477 4% 2944, 3073
On Thu, Apr 7, 2011 at 8:29 PM, Simon Plouffe <simon.plouffe@gmail.com wrote:
Hello,
this is the finite differences of Newton,
n^3 = (1,7,12,6) n^2 = (1,3,2) n = (1,1) 1 = (1)
then if we have (37,46,28,6) as coefficients of the differences, a linear combination computation gives ; n^3 + 8*n^2 + 15*n + 13
37, 83, 157, 265, 413, 607, 853, 1157, 1525, 1963, 2477, 3073, ...
the coefficients of n^k are related to Stirling numbers.
ref : The book of Jordan < calculus of finite differences>.
http://books.google.com/books?id=3RfZOsDAyQsC&printsec=frontcover&dq=finite+...
best regards,
Simon Plouffe
Le 08/avr./2011 05:09, Joshua Zucker a écrit :
My method (use fixed-width font for viewing):
37 83 157 ? 46 74 28 6
Has to be a 6 down there, to make the leading coefficient 1.
Thus:
37 83 157 265 46 74 108 28 34 6
Answer: 265.
--Joshua _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Check out Golly at http://golly.sf.net/
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No, I definitely agree. But there's something amusing about the intersection of logic and math here, the fact that we can find a mathematical solution that ignores the physical realities, but because the problem is posed as a physical problem so its preconditions are impossible, logic dictates that any answer is as acceptable as the "correct" one. On Fri, Apr 8, 2011 at 4:53 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Tom — I think the original problem is interesting by itself. I just mean that if you graph the following there's no place where all three surfaces intersect. xyz=37 (x+1)(y+1)(z+1)=83 (x+2)(y+2)(z+2)=157
On Fri, Apr 8, 2011 at 4:30 PM, Tom Rokicki <rokicki@gmail.com> wrote:
So 42 cubic inches is a correct answer to the original question, correct? Or at least as valid as the other answers?
David, would you mark "42 cubic inches" as correct?
This reminds me of the little-known fact that if you hold up a guinea pig by its tail, its eyes fall out (although strangely pet stores seldom warn of this).
-tom
On Fri, Apr 8, 2011 at 4:12 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Ah - probably not your point, David, but this can't be a real box. No box would ever have volumes 37, 83, 156 as you increase the length of each side by 1. (At least one that's not non-imaginary.)
You can get close to your series — volumes 36, 83, 156, 263 — starting with a box 4.25 by 4.25 by 2, for example.
On Fri, Apr 8, 2011 at 12:10 AM, Gary Antonick <gantonick@post.harvard.edu>wrote:
Seems you could use the first two numbers to determine you have essentially a cube: (37^(1/3)+1)^3 = 81.3, which is pretty close to 83. Next two boxes are predicted 152 and 254, actual 157 and 265.
Interesting that the error climbs to 5% then falls as the box gets more and more cube-like.
2% 81, 83 4% 152, 157 4% 254, 265 5% 394, 413 5% 578, 607 5% 813, 853 5% 1103, 1157 5% 1455, 1525 5% 1876, 1963 5% 2370, 2477 4% 2944, 3073
On Thu, Apr 7, 2011 at 8:29 PM, Simon Plouffe <simon.plouffe@gmail.com wrote:
Hello,
this is the finite differences of Newton,
n^3 = (1,7,12,6) n^2 = (1,3,2) n = (1,1) 1 = (1)
then if we have (37,46,28,6) as coefficients of the differences, a linear combination computation gives ; n^3 + 8*n^2 + 15*n + 13
37, 83, 157, 265, 413, 607, 853, 1157, 1525, 1963, 2477, 3073, ...
the coefficients of n^k are related to Stirling numbers.
ref : The book of Jordan < calculus of finite differences>.
http://books.google.com/books?id=3RfZOsDAyQsC&printsec=frontcover&dq=finite+...
best regards,
Simon Plouffe
Le 08/avr./2011 05:09, Joshua Zucker a écrit :
My method (use fixed-width font for viewing):
37 83 157 ? 46 74 28 6
Has to be a 6 down there, to make the leading coefficient 1.
Thus:
37 83 157 265 46 74 108 28 34 6
Answer: 265.
--Joshua _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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ah. yes. or -3, or 3i. What's interesting is that the dimensions sound good at first. Like a triangle with sides 24, 8, and 15. The numbers are close, though, so am wondering if some reasonable situation the question might make sense. A bad ruler, for example, or bad eyesight. A misshapen box. The series is really 37-ish, 83-ish, etc.. in which case it could be argued that a reasonable response might be something between 260 and 270. Just right for several guinea pigs. otherwise probably best kept aside for unicorns. am wondering what David had in mind. the impossibility gives this a nice twist. On Fri, Apr 8, 2011 at 5:00 PM, Tom Rokicki <rokicki@gmail.com> wrote:
No, I definitely agree. But there's something amusing about the intersection of logic and math here, the fact that we can find a mathematical solution that ignores the physical realities, but because the problem is posed as a physical problem so its preconditions are impossible, logic dictates that any answer is as acceptable as the "correct" one.
On Fri, Apr 8, 2011 at 4:53 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Tom — I think the original problem is interesting by itself. I just mean that if you graph the following there's no place where all three surfaces intersect. xyz=37 (x+1)(y+1)(z+1)=83 (x+2)(y+2)(z+2)=157
On Fri, Apr 8, 2011 at 4:30 PM, Tom Rokicki <rokicki@gmail.com> wrote:
So 42 cubic inches is a correct answer to the original question, correct? Or at least as valid as the other answers?
David, would you mark "42 cubic inches" as correct?
This reminds me of the little-known fact that if you hold up a guinea pig by its tail, its eyes fall out (although strangely pet stores seldom warn of this).
-tom
On Fri, Apr 8, 2011 at 4:12 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Ah - probably not your point, David, but this can't be a real box. No box would ever have volumes 37, 83, 156 as you increase the length of each side by 1. (At least one that's not non-imaginary.)
You can get close to your series — volumes 36, 83, 156, 263 — starting with a box 4.25 by 4.25 by 2, for example.
On Fri, Apr 8, 2011 at 12:10 AM, Gary Antonick <gantonick@post.harvard.edu>wrote:
Seems you could use the first two numbers to determine you have essentially a cube: (37^(1/3)+1)^3 = 81.3, which is pretty close to 83. Next two boxes are predicted 152 and 254, actual 157 and 265.
Interesting that the error climbs to 5% then falls as the box gets more and more cube-like.
2% 81, 83 4% 152, 157 4% 254, 265 5% 394, 413 5% 578, 607 5% 813, 853 5% 1103, 1157 5% 1455, 1525 5% 1876, 1963 5% 2370, 2477 4% 2944, 3073
On Thu, Apr 7, 2011 at 8:29 PM, Simon Plouffe < simon.plouffe@gmail.com wrote:
Hello,
this is the finite differences of Newton,
n^3 = (1,7,12,6) n^2 = (1,3,2) n = (1,1) 1 = (1)
then if we have (37,46,28,6) as coefficients of the differences, a linear combination computation gives ; n^3 + 8*n^2 + 15*n + 13
37, 83, 157, 265, 413, 607, 853, 1157, 1525, 1963, 2477, 3073, ...
the coefficients of n^k are related to Stirling numbers.
ref : The book of Jordan < calculus of finite differences>.
http://books.google.com/books?id=3RfZOsDAyQsC&printsec=frontcover&dq=finite+...
best regards,
Simon Plouffe
Le 08/avr./2011 05:09, Joshua Zucker a écrit :
My method (use fixed-width font for viewing): > > 37 83 157 ? > 46 74 > 28 > 6 > > Has to be a 6 down there, to make the leading coefficient 1. > > Thus: > > 37 83 157 265 > 46 74 108 > 28 34 > 6 > > Answer: 265. > > > --Joshua > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Don't know what to tell you on this one. My original intention was to specify an actual box with real sides. I started with a 2.2 x 3.2 x 5.2 box, and rounded the volumes to integers, assuming there would be a nearby box with similar volumes. To my chagrin, tweaking the volumes to integers has apparently driven the sides to complex land. I was too clever (read lazy) for my own good, but Simon Plouffe can take solace that I fooled myself as well. About the 42... Under my original supposition of a R^3 box, there are no (positive) real box dimensions satisfying the problem. So 42 gets marked wrong. 42 is indeed as valid as any other numerical answer, that is to say, incorrect, but it is not as valid as the answer "no solution." For a C^3 box, 265 is the only solution. 42 again gets marked wrong. If the box were in C^4, 42 would squeak by, but I specifically wrote "cubic inches", not "tesseractal inches", so no dice. I wish people would just do their homework and not try to slide by without working (like me). On 4/8/2011 7:30 PM, Tom Rokicki wrote:
So 42 cubic inches is a correct answer to the original question, correct? Or at least as valid as the other answers?
David, would you mark "42 cubic inches" as correct?
This reminds me of the little-known fact that if you hold up a guinea pig by its tail, its eyes fall out (although strangely pet stores seldom warn of this).
-tom
On Fri, Apr 8, 2011 at 4:12 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
participants (5)
-
David Wilson -
Gary Antonick -
Joshua Zucker -
Simon Plouffe -
Tom Rokicki