Question #d5a5c

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Answer 1

May 5, 2016

#x=3#

Note that as #4-=-1" (mod 5)"# we have

#2^2010 -=(2^2)^1005 -= 4^(1005) -= (-1)^1005 -= -1" (mod 5)"#

Thus we are actually searching for the least #x inZZ^+# such that

#3x -= -1" (mod 5)"#

From here, we can simply iterate #x# starting from #x=1#.

#3(1) -= 3 cancel(-=) -1" (mod 5)"#

#3(2) -= 6 -= 1 cancel(-=) -1" (mod 5)"#

#3(3) -= 9 -= -1" (mod 5)"#

Therefore the least such positive integer #x# is #x=3#