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 Move the 2 - double the number II (Posted on 2011-08-21)
(I) A certain tridecimal (base 13) positive integer starts with the digit 2. Moving the 2 from the beginning of the number to its end doubles it.

What is the minimum value of this number?

(II) Determine the general form of a positive integer n such that there does not exist any base-n positive integer, starting with the digit 2, that doubles itself when the digit 2 is shifted from the beginning of the number to its end.

 No Solution Yet Submitted by K Sengupta No Rating

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 Part I solution | Comment 1 of 2

In an (n+1)-digit number:

2*(2*13^n + x) = 13*x + 2

4*13^n + 2*x = 13*x + 2

11*x = 4*13^n - 2

x = (4*13^n - 2) / 11

for some values of n:

` n         x 0       2/11 1       50/11 2       674/11 3       8786/11 4       114242/11 5       1485170/11 6       19307234/11 7       250994066/11 8       3262922882/11 9       3856181590 10      551433967394/11 11      7168641576146/11 12      93192340489922/11 13      1211500426369010/11 14      15749505542797154/11 15      204743572056363026/11 16      2661666436732719362/11 17      34601663677525351730/11 18      449821627807829572514/11 19      531607378318344040246 20      76019855099523197755202/11 21      988258116293801570817650/11 `

so the first n that works (x is an integer) is 9.

The 3,856,181,590 must be converted to base-13 before we can put a 2 in front of it and call part I solved. That conversion comes out to 495BA8371, so the answer to part I is:

2495BA8371

 Posted by Charlie on 2011-08-21 13:47:32

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