Show that the remainder when 2^1990 (2 to the power of 1990) is divided by 1990 equals 1024.
As 1990=2*5*199, 2^1990 can be written as (2^10)^199, or 1024^199, so 1024^199 has to be shown congruent to 1024 mod 1990.
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Posted by Charlie
on 2003-03-02 09:16:10 |