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