I came up with an answer in about 10 seconds, but I think it's cheating - it's at least cheap. It seems like if there's a real answer a little algebra should uncover it. To the scratch paper!
I think the answer is that there can be no such integer (aside from the cheap one). According to my scratchings, in order for there to be such an integer, there must be some n > 0 such that 19 divides (10^n) - 2. A common reference site suggests that to test divisibility by 19, you do the following:
--chop off the last digit of the number in question, double it, and add it to the "rest" of the number --repeat until you get to either a known multiple or a known non-multiple of 19.
According to some more noodling, you'll never get to a multiple of 19 starting from (10^n) - 2, for n in the positive integers.
The cheap answer must be the additive identity. While scratching I also concluded that there could be no such integer, but then changed my mind, finally arriving at the 10^n = 19m + 2 expression. I turned to Excel instead of looking online. The MOD function is buggy, failing for MOD(N,19) when N>2550136831. The equivalent N-(19*INT(N/19)) works better, but reports false positives for candidates beyond 999999999999998 due to rounding error.
Windows' calc.exe indicates that (10^19 - 2) % 19 = 8. However, 10^17 - 2 seems to work, but this is only seventeen digits long. A simple script also finds this the smallest solution. Who is wrong -- the reporter? Me? Rasalom? Surely not Dyson?
Ah, I was wrong - punched the wrong numbers into calc, I think.
But before suspecting FD of error, remember that 10^17 - 2 is not the solution to the problem; it's only the left side of the equation along with D0. D1, D2, ..., D17 (where Dj is the jth digit of the solution) are on the right side, for a total of 18 digits.
Oh, of course; I lost sight of the big picture. The best (-intentioned) arithmeticians make the worst mathematicians. Now I've lost my scratch paper and can't even work out what the lowest solution is.
A cleaner script shows a pattern indicating multiple solutions clearly.
So how did Dyson do it? Perhaps he had considered the problem before, or some similar problem. Perhaps the "two seconds" was an exaggeration, and he worked out the algebra and applied the divisibility by 19 rule in some fraction of a minute which amazed his company and distorted their sense of time.
What's that quote about a maths expert being someone who can do something you could do if you just worked long enough, but a genius being [something else]?
It was Mark Kac, speaking of Richard Feynman as a magician rather than an ordinary genius:
An ordinary genius is a fellow that you and I would be just as good as, if we were only many times better. There is no mystery as to how his mind works. Once we understand what they have done, we feel certain that we, too, could have done it. It is different with the magicians... Even after we understand what they have done, the process by which they have done it is completely dark.
God, I miss Feynman. Isn't that weird? I'd barely heard of him when he died. Here's a gem from the wikipedia article - an excerpt from his second wife's divorce complaint:
He begins working calculus problems in his head as soon as he awakens. He did calculus while driving in his car, while sitting in the living room, and while lying in bed at night. —Mary Louise Bell divorce complaint[40]p.168
I just freakin' love that guy. Fuck you, death. Fuck you dead with a salted pineapple.
13 comments:
I came up with an answer in about 10 seconds, but I think it's cheating - it's at least cheap. It seems like if there's a real answer a little algebra should uncover it. To the scratch paper!
I think the answer is that there can be no such integer (aside from the cheap one). According to my scratchings, in order for there to be such an integer, there must be some n > 0 such that 19 divides (10^n) - 2. A common reference site suggests that to test divisibility by 19, you do the following:
--chop off the last digit of the number in question, double it, and add it to the "rest" of the number
--repeat until you get to either a known multiple or a known non-multiple of 19.
According to some more noodling, you'll never get to a multiple of 19 starting from (10^n) - 2, for n in the positive integers.
Now to check the spoiler...
Dang. I stand by my pre-noodling analysis, though.
Ha ha, and in fact 10^19 - 2 is divisible by 19. Freeman, you old rascal.
The cheap answer must be the additive identity. While scratching I also concluded that there could be no such integer, but then changed my mind, finally arriving at the 10^n = 19m + 2 expression. I turned to Excel instead of looking online. The MOD function is buggy, failing for MOD(N,19) when N>2550136831. The equivalent N-(19*INT(N/19)) works better, but reports false positives for candidates beyond 999999999999998 due to rounding error.
Windows' calc.exe indicates that (10^19 - 2) % 19 = 8. However, 10^17 - 2 seems to work, but this is only seventeen digits long. A simple script also finds this the smallest solution. Who is wrong -- the reporter? Me? Rasalom? Surely not Dyson?
Ah, I was wrong - punched the wrong numbers into calc, I think.
But before suspecting FD of error, remember that 10^17 - 2 is not the solution to the problem; it's only the left side of the equation along with D0. D1, D2, ..., D17 (where Dj is the jth digit of the solution) are on the right side, for a total of 18 digits.
Freeman, you scamp! You rapscallion!
Oops, I meant Dj is the (j minus 1)th digit. I.e. the coefficient of 10^j in the expansion. You know what I meant.
(j plus 1)th.
Oh, of course; I lost sight of the big picture. The best (-intentioned) arithmeticians make the worst mathematicians. Now I've lost my scratch paper and can't even work out what the lowest solution is.
A cleaner script shows a pattern indicating multiple solutions clearly.
So how did Dyson do it? Perhaps he had considered the problem before, or some similar problem. Perhaps the "two seconds" was an exaggeration, and he worked out the algebra and applied the divisibility by 19 rule in some fraction of a minute which amazed his company and distorted their sense of time.
What's that quote about a maths expert being someone who can do something you could do if you just worked long enough, but a genius being [something else]?
It was Mark Kac, speaking of Richard Feynman as a magician rather than an ordinary genius:
An ordinary genius is a fellow that you and I would be just as good as, if we were only many times better. There is no mystery as to how his mind works. Once we understand what they have done, we feel certain that we, too, could have done it. It is different with the magicians... Even after we understand what they have done, the process by which they have done it is completely dark.
God, I miss Feynman. Isn't that weird? I'd barely heard of him when he died. Here's a gem from the wikipedia article - an excerpt from his second wife's divorce complaint:
He begins working calculus problems in his head as soon as he awakens. He did calculus while driving in his car, while sitting in the living room, and while lying in bed at night.
—Mary Louise Bell divorce complaint[40]p.168
I just freakin' love that guy. Fuck you, death. Fuck you dead with a salted pineapple.
<del>pineapple</del>durian
<del>durian</del>jackfruit
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