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Maximum Power 2 Divide Evenly (Posted on 2008-10-19) |
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Determine the maximum value of a positive integer G, such that (20071024 -1) is evenly divisible by 2G.
Note: Try to solve this problem analytically, although computer program/ spreadsheet solutions are welcome.
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Submitted by K Sengupta
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Rating: 4.0000 (2 votes)
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Solution:
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The required maximum value of maximum value of G is 13.
EXPLANATION:
Factorizing the given expression, we have:
(2007^1024 – 1) = (2007^512 + 1) (2007^256 + 1) (2007^128 + 1) (2007^64+ 1)……..*(2007^2 + 1)*(2007+1)(2007-1) ……(i)
With the exception of (2007-1), all the other factors possess the form (20072^m +1).
For m >= 1, we observe that:
20072^m(mod 4) = (-1)2^m (mod 4) = 1
or, (20072^m+1) (mod 4) = 2
Thus, each the first 9 factors in (i) are divisible by 2, and no higher power of 2 divides these factors.
For the two remaining factors, we observe that the highest power of 2 dividing (2007-1) = 2006 is 1 and the highest power of 2 dividing (2007+1) = 2008 is 3.
Consequently, the required maximum value of maximum value of G is: (9+1+3) = 13.
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