[math-fun] Freeman Dyson integer problem
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html "At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, its possible to exactly double the value. Dyson will immediately say, Oh, thats not difficult, allow two short beats to pass and then add, but of course the smallest such number is 18 digits long. When this happened one day at lunch, William Press remembers, the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds. The meal then ended with men who tend to be described with words like brilliant, Nobel and MacArthur quietly retreating to their offices to work out what Dyson just knew." Is this correct?
421052631578947368 / 210526315789473684 = 2 smallest k, such that 10^k - 2 is divisible by 2 x 10 - 1 = 19, is k = 17 Veit On Mar 26, 2009, at 3:52 PM, Henry Baker wrote:
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html
"At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, its possible to exactly double the value. Dyson will immediately say, Oh, thats not difficult, allow two short beats to pass and then add, but of course the smallest such number is 18 digits long. When this happened one day at lunch, William Press remembers, the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds. The meal then ended with men who tend to be described with words like brilliant, Nobel and MacArthur quietly retreating to their offices to work out what Dyson just knew."
Is this correct?
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So, one can just start with a digit and start working backwards, at each point dividing by 2 and seeing if it "works" when the digit repeats? E.g., 315789473684210526 will also work; 421052631578947368, etc. Bill C. -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Veit Elser Sent: Thursday, March 26, 2009 9:17 AM To: math-fun Subject: Re: [math-fun] Freeman Dyson integer problem 421052631578947368 / 210526315789473684 = 2 smallest k, such that 10^k - 2 is divisible by 2 x 10 - 1 = 19, is k = 17 Veit On Mar 26, 2009, at 3:52 PM, Henry Baker wrote:
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html
"At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it's possible to exactly double the value. Dyson will immediately say, "Oh, that's not difficult," allow two short beats to pass and then add, "but of course the smallest such number is 18 digits long." When this happened one day at lunch, William Press remembers, "the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds." The meal then ended with men who tend to be described with words like "brilliant," "Nobel" and "MacArthur" quietly retreating to their offices to work out what Dyson just knew."
Is this correct?
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For some reason I haven't quite figured out the algebra, but permit me to report that 923076/3 = 307692, with two digits moving to the back. On Thu, Mar 26, 2009 at 12:12 PM, Cordwell, William R <wrcordw@sandia.gov>wrote:
So, one can just start with a digit and start working backwards, at each point dividing by 2 and seeing if it "works" when the digit repeats? E.g., 315789473684210526 will also work; 421052631578947368, etc.
Bill C.
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto: math-fun-bounces@mailman.xmission.com] On Behalf Of Veit Elser Sent: Thursday, March 26, 2009 9:17 AM To: math-fun Subject: Re: [math-fun] Freeman Dyson integer problem
421052631578947368 / 210526315789473684 = 2
smallest k, such that 10^k - 2 is divisible by 2 x 10 - 1 = 19, is k = 17
Veit
On Mar 26, 2009, at 3:52 PM, Henry Baker wrote:
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html
"At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it's possible to exactly double the value. Dyson will immediately say, "Oh, that's not difficult," allow two short beats to pass and then add, "but of course the smallest such number is 18 digits long." When this happened one day at lunch, William Press remembers, "the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds." The meal then ended with men who tend to be described with words like "brilliant," "Nobel" and "MacArthur" quietly retreating to their offices to work out what Dyson just knew."
Is this correct?
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A few things about this. My technique also works for base 100, which is what you are doing here. X/100 + d/100 = 3X X = d/299 for d=92, we get X=92/299=4/13, and since the order of 10 mod 13 is 6, we get a six digit answer. 307692 is the repeating block of the decimal expansion of 4/13. For a smaller answer, try the repeating block of 2/13. 153846*3=461538. On Thu, Mar 26, 2009 at 10:15 AM, Allan Wechsler <acwacw@gmail.com> wrote:
For some reason I haven't quite figured out the algebra, but permit me to report that 923076/3 = 307692, with two digits moving to the back.
On Thu, Mar 26, 2009 at 12:12 PM, Cordwell, William R <wrcordw@sandia.gov
wrote:
So, one can just start with a digit and start working backwards, at each point dividing by 2 and seeing if it "works" when the digit repeats? E.g., 315789473684210526 will also work; 421052631578947368, etc.
Bill C.
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto: math-fun-bounces@mailman.xmission.com] On Behalf Of Veit Elser Sent: Thursday, March 26, 2009 9:17 AM To: math-fun Subject: Re: [math-fun] Freeman Dyson integer problem
421052631578947368 / 210526315789473684 = 2
smallest k, such that 10^k - 2 is divisible by 2 x 10 - 1 = 19, is k = 17
Veit
On Mar 26, 2009, at 3:52 PM, Henry Baker wrote:
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html
"At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it's possible to exactly double the value. Dyson will immediately say, "Oh, that's not difficult," allow two short beats to pass and then add, "but of course the smallest such number is 18 digits long." When this happened one day at lunch, William Press remembers, "the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds." The meal then ended with men who tend to be described with words like "brilliant," "Nobel" and "MacArthur" quietly retreating to their offices to work out what Dyson just knew."
Is this correct?
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That's exactly the technique I was trying to apply for base 10. For some reason I slipped and solved 26S = 3a (where a is the leading digit) rather than 29S = 3a, so I wound up with a base-100 solution. On Thu, Mar 26, 2009 at 1:35 PM, Dave Blackston <hyperdex@gmail.com> wrote:
A few things about this.
My technique also works for base 100, which is what you are doing here.
X/100 + d/100 = 3X
X = d/299
for d=92, we get X=92/299=4/13, and since the order of 10 mod 13 is 6, we get a six digit answer.
307692 is the repeating block of the decimal expansion of 4/13.
For a smaller answer, try the repeating block of 2/13. 153846*3=461538.
On Thu, Mar 26, 2009 at 10:15 AM, Allan Wechsler <acwacw@gmail.com> wrote:
For some reason I haven't quite figured out the algebra, but permit me to report that 923076/3 = 307692, with two digits moving to the back.
On Thu, Mar 26, 2009 at 12:12 PM, Cordwell, William R < wrcordw@sandia.gov
wrote:
So, one can just start with a digit and start working backwards, at each point dividing by 2 and seeing if it "works" when the digit repeats? E.g., 315789473684210526 will also work; 421052631578947368, etc.
Bill C.
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto: math-fun-bounces@mailman.xmission.com] On Behalf Of Veit Elser Sent: Thursday, March 26, 2009 9:17 AM To: math-fun Subject: Re: [math-fun] Freeman Dyson integer problem
421052631578947368 / 210526315789473684 = 2
smallest k, such that 10^k - 2 is divisible by 2 x 10 - 1 = 19, is k = 17
Veit
On Mar 26, 2009, at 3:52 PM, Henry Baker wrote:
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html
"At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it's possible to exactly double the value. Dyson will immediately say, "Oh, that's not difficult," allow two short beats to pass and then add, "but of course the smallest such number is 18 digits long." When this happened one day at lunch, William Press remembers, "the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds." The meal then ended with men who tend to be described with words like "brilliant," "Nobel" and "MacArthur" quietly retreating to their offices to work out what Dyson just knew."
Is this correct?
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Using the same idea, start with 2, and consecutively divide by 3, to obtain 3*2068965517241379310345827586 = 6206896551724137931037482758 -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Thursday, March 26, 2009 11:16 AM To: math-fun Subject: Re: [math-fun] Freeman Dyson integer problem For some reason I haven't quite figured out the algebra, but permit me to report that 923076/3 = 307692, with two digits moving to the back. On Thu, Mar 26, 2009 at 12:12 PM, Cordwell, William R <wrcordw@sandia.gov>wrote:
So, one can just start with a digit and start working backwards, at each point dividing by 2 and seeing if it "works" when the digit repeats? E.g., 315789473684210526 will also work; 421052631578947368, etc.
Bill C.
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto: math-fun-bounces@mailman.xmission.com] On Behalf Of Veit Elser Sent: Thursday, March 26, 2009 9:17 AM To: math-fun Subject: Re: [math-fun] Freeman Dyson integer problem
421052631578947368 / 210526315789473684 = 2
smallest k, such that 10^k - 2 is divisible by 2 x 10 - 1 = 19, is k = 17
Veit
On Mar 26, 2009, at 3:52 PM, Henry Baker wrote:
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html
"At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it's possible to exactly double the value. Dyson will immediately say, "Oh, that's not difficult," allow two short beats to pass and then add, "but of course the smallest such number is 18 digits long." When this happened one day at lunch, William Press remembers, "the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds." The meal then ended with men who tend to be described with words like "brilliant," "Nobel" and "MacArthur" quietly retreating to their offices to work out what Dyson just knew."
Is this correct?
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Good. That's a rotation of the one I posted earlier, which started 3*10374... On Thu, Mar 26, 2009 at 1:56 PM, Cordwell, William R <wrcordw@sandia.gov>wrote:
Using the same idea, start with 2, and consecutively divide by 3, to obtain 3*2068965517241379310345827586 = 6206896551724137931037482758
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto: math-fun-bounces@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Thursday, March 26, 2009 11:16 AM To: math-fun Subject: Re: [math-fun] Freeman Dyson integer problem
For some reason I haven't quite figured out the algebra, but permit me to report that 923076/3 = 307692, with two digits moving to the back.
On Thu, Mar 26, 2009 at 12:12 PM, Cordwell, William R <wrcordw@sandia.gov
wrote:
So, one can just start with a digit and start working backwards, at each point dividing by 2 and seeing if it "works" when the digit repeats? E.g., 315789473684210526 will also work; 421052631578947368, etc.
Bill C.
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto: math-fun-bounces@mailman.xmission.com] On Behalf Of Veit Elser Sent: Thursday, March 26, 2009 9:17 AM To: math-fun Subject: Re: [math-fun] Freeman Dyson integer problem
421052631578947368 / 210526315789473684 = 2
smallest k, such that 10^k - 2 is divisible by 2 x 10 - 1 = 19, is k = 17
Veit
On Mar 26, 2009, at 3:52 PM, Henry Baker wrote:
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html
"At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it's possible to exactly double the value. Dyson will immediately say, "Oh, that's not difficult," allow two short beats to pass and then add, "but of course the smallest such number is 18 digits long." When this happened one day at lunch, William Press remembers, "the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds." The meal then ended with men who tend to be described with words like "brilliant," "Nobel" and "MacArthur" quietly retreating to their offices to work out what Dyson just knew."
Is this correct?
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Suppose that the number in question is 10n + a (where a is the lowest digit). The transformed number is a*10^r + n. If that's equal to double the original we get a(10^r - 2) = 19 n, so that 19 has to divide 10^r-2 (since it's prime and bigger than a). 10 is a primitive root mod 19, and you can check that 2 = 10^17 mod 19, so that r must be 17 mod 18. You can check that (10^17 - 2)/19 = 2*q, where q is a prime. Thus the number in question is 10*q + 2. Victor ---------- Forwarded message ---------- From: Henry Baker <hbaker1@pipeline.com> Date: Thu, Mar 26, 2009 at 10:52 AM Subject: [math-fun] Freeman Dyson integer problem To: math-fun@mailman.xmission.com http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html "At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it’s possible to exactly double the value. Dyson will immediately say, “Oh, that’s not difficult,” allow two short beats to pass and then add, “but of course the smallest such number is 18 digits long.” When this happened one day at lunch, William Press remembers, “the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds.” The meal then ended with men who tend to be described with words like “brilliant,” “Nobel” and “MacArthur” quietly retreating to their offices to work out what Dyson just knew." Is this correct? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
The explanation is that Freeman Dyson (who started his career as a Number Theorist in case you've forgotten) remembered that 10 was a primitive root mod 19, and observed that 2*10 = 1 mod 19, so that 2 would have to be a "tail end" of the powers of 10 mod 19. Victor On Thu, Mar 26, 2009 at 11:29 AM, victor miller <victorsmiller@gmail.com>wrote:
Suppose that the number in question is 10n + a (where a is the lowest digit). The transformed number is a*10^r + n. If that's equal to double the original we get
a(10^r - 2) = 19 n, so that 19 has to divide 10^r-2 (since it's prime and bigger than a). 10 is a primitive root mod 19, and you can check that 2 = 10^17 mod 19, so that r must be 17 mod 18. You can check that (10^17 - 2)/19 = 2*q, where q is a prime. Thus the number in question is 10*q + 2.
Victor
---------- Forwarded message ---------- From: Henry Baker <hbaker1@pipeline.com> Date: Thu, Mar 26, 2009 at 10:52 AM Subject: [math-fun] Freeman Dyson integer problem To: math-fun@mailman.xmission.com
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html
"At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it’s possible to exactly double the value. Dyson will immediately say, “Oh, that’s not difficult,” allow two short beats to pass and then add, “but of course the smallest such number is 18 digits long.” When this happened one day at lunch, William Press remembers, “the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds.” The meal then ended with men who tend to be described with words like “brilliant,” “Nobel” and “MacArthur” quietly retreating to their offices to work out what Dyson just knew."
Is this correct?
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A simpler explanation is that he's actually done this problem before and remembered the answer. I'm no Freeman Dyson, but I've seen (and solved) this problem before. In fact I believe a variant of this problem appeared on this list and was forwarded to me by a friend, which is how I ended up here. My solution is a little different... suppose x is our number, and it has last digit d. Further, let X be the rational number formed by repeating x after a decimal point, i.e., if x is 112, then X is .112112112112... The constraints imply that X/10 + d/10 = 2X, or X=d/19. Clearly X>.1, so d>1. This means the smallest such X occurs for d=2, X=2/19, and x is the repeated block for the decimal 2/19, which has 18 digits since 10 is a primitive root mod 19. On Thu, Mar 26, 2009 at 8:32 AM, victor miller <victorsmiller@gmail.com>wrote:
The explanation is that Freeman Dyson (who started his career as a Number Theorist in case you've forgotten) remembered that 10 was a primitive root mod 19, and observed that 2*10 = 1 mod 19, so that 2 would have to be a "tail end" of the powers of 10 mod 19.
Victor
On Thu, Mar 26, 2009 at 11:29 AM, victor miller <victorsmiller@gmail.com
wrote:
Suppose that the number in question is 10n + a (where a is the lowest digit). The transformed number is a*10^r + n. If that's equal to double the original we get
a(10^r - 2) = 19 n, so that 19 has to divide 10^r-2 (since it's prime and bigger than a). 10 is a primitive root mod 19, and you can check that 2 = 10^17 mod 19, so that r must be 17 mod 18. You can check that (10^17 - 2)/19 = 2*q, where q is a prime. Thus the number in question is 10*q + 2.
Victor
---------- Forwarded message ---------- From: Henry Baker <hbaker1@pipeline.com> Date: Thu, Mar 26, 2009 at 10:52 AM Subject: [math-fun] Freeman Dyson integer problem To: math-fun@mailman.xmission.com
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html
"At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it’s possible to exactly double the value. Dyson will immediately say, “Oh, that’s not difficult,” allow two short beats to pass and then add, “but of course the smallest such number is 18 digits long.” When this happened one day at lunch, William Press remembers, “the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds.” The meal then ended with men who tend to be described with words like “brilliant,” “Nobel” and “MacArthur” quietly retreating to their offices to work out what Dyson just knew."
Is this correct?
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This problem appears on page 76 of Maurice Kraitchik's Mathematical Recreations, 2nd edition, Dover 1953. The first edition was in 1941 (Norton), and the preface mentions it's an upgrade of his French work from 1930, with augmentation from several sources including Sphinx. Kraitchik is underappreciated as a mathematician, but it would not be at all surprising if Dyson had read his book. I've seen the 19 puzzle in at least one other place, and the simpler 7 version in several puzzle books. Parlo(u)r Trick indeed! Rich ---------------- Quoting victor miller <victorsmiller@gmail.com>:
The explanation is that Freeman Dyson (who started his career as a Number Theorist in case you've forgotten) remembered that 10 was a primitive root mod 19, and observed that 2*10 = 1 mod 19, so that 2 would have to be a "tail end" of the powers of 10 mod 19.
Victor
On Thu, Mar 26, 2009 at 11:29 AM, victor miller <victorsmiller@gmail.com>wrote:
Suppose that the number in question is 10n + a (where a is the lowest digit). The transformed number is a*10^r + n. If that's equal to double the original we get
a(10^r - 2) = 19 n, so that 19 has to divide 10^r-2 (since it's prime and bigger than a). 10 is a primitive root mod 19, and you can check that 2 = 10^17 mod 19, so that r must be 17 mod 18. You can check that (10^17 - 2)/19 = 2*q, where q is a prime. Thus the number in question is 10*q + 2.
Victor
---------- Forwarded message ---------- From: Henry Baker <hbaker1@pipeline.com> Date: Thu, Mar 26, 2009 at 10:52 AM Subject: [math-fun] Freeman Dyson integer problem To: math-fun@mailman.xmission.com
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html
"At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it’s possible to exactly double the value. Dyson will immediately say, “Oh, that’s not difficult,” allow two short beats to pass and then add, “but of course the smallest such number is 18 digits long.” When this happened one day at lunch, William Press remembers, “the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds.” The meal then ended with men who tend to be described with words like “brilliant,” “Nobel” and “MacArthur” quietly retreating to their offices to work out what Dyson just knew."
Is this correct?
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I have Kraitchik somewhere. It's an underappreciated gem. On Thu, Mar 26, 2009 at 1:32 PM, <rcs@xmission.com> wrote:
This problem appears on page 76 of Maurice Kraitchik's Mathematical Recreations, 2nd edition, Dover 1953. The first edition was in 1941 (Norton), and the preface mentions it's an upgrade of his French work from 1930, with augmentation from several sources including Sphinx. Kraitchik is underappreciated as a mathematician, but it would not be at all surprising if Dyson had read his book. I've seen the 19 puzzle in at least one other place, and the simpler 7 version in several puzzle books. Parlo(u)r Trick indeed!
Rich
----------------
Quoting victor miller <victorsmiller@gmail.com>:
The explanation is that Freeman Dyson (who started his career as a Number
Theorist in case you've forgotten) remembered that 10 was a primitive root mod 19, and observed that 2*10 = 1 mod 19, so that 2 would have to be a "tail end" of the powers of 10 mod 19.
Victor
On Thu, Mar 26, 2009 at 11:29 AM, victor miller <victorsmiller@gmail.com
wrote:
Suppose that the number in question is 10n + a (where a is the lowest
digit). The transformed number is a*10^r + n. If that's equal to double the original we get
a(10^r - 2) = 19 n, so that 19 has to divide 10^r-2 (since it's prime and bigger than a). 10 is a primitive root mod 19, and you can check that 2 = 10^17 mod 19, so that r must be 17 mod 18. You can check that (10^17 - 2)/19 = 2*q, where q is a prime. Thus the number in question is 10*q + 2.
Victor
---------- Forwarded message ---------- From: Henry Baker <hbaker1@pipeline.com> Date: Thu, Mar 26, 2009 at 10:52 AM Subject: [math-fun] Freeman Dyson integer problem To: math-fun@mailman.xmission.com
http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html
"At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it’s possible to exactly double the value. Dyson will immediately say, “Oh, that’s not difficult,” allow two short beats to pass and then add, “but of course the smallest such number is 18 digits long.” When this happened one day at lunch, William Press remembers, “the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds.” The meal then ended with men who tend to be described with words like “brilliant,” “Nobel” and “MacArthur” quietly retreating to their offices to work out what Dyson just knew."
Is this correct?
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Sounds like a legend, or like something Von Neumann or Ramanujan might do. I don't think Dyson, as brilliant as he is, was noted for amazing mental calculations. I could be wrong.* * Although of course it's never happened before! :) -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Henry Baker Sent: Thursday, March 26, 2009 7:52 AM To: math-fun@mailman.xmission.com Subject: [math-fun] Freeman Dyson integer problem http://www.nytimes.com/2009/03/29/magazine/29Dyson-t.html "At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, its possible to exactly double the value. Dyson will immediately say, Oh, thats not difficult, allow two short beats to pass and then add, but of course the smallest such number is 18 digits long. When this happened one day at lunch, William Press remembers, the table fell silent; nobody had the slightest idea how Freeman could have known such a fact or, even more terrifying, could have derived it in his head in about two seconds. The meal then ended with men who tend to be described with words like brilliant, Nobel and MacArthur quietly retreating to their offices to work out what Dyson just knew." Is this correct?
participants (8)
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Allan Wechsler -
Cordwell, William R -
Dave Blackston -
Henry Baker -
rcs＠xmission.com -
Stephen Gray -
Veit Elser -
victor miller