Question #d5a5c

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Question #d5a5c

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Answer 1 · 2016-05-05T08:35:40

$x = 3$ $x=3$

Explanation:

Note that as $4 \equiv - 1 (mod 5)$ $4-=-1" (mod 5)"$ we have

$2^{2010} \equiv {(2^{2})}^{1005} \equiv 4^{1005} \equiv {(- 1)}^{1005} \equiv - 1 (mod 5)$ $2^2010 -=(2^2)^1005 -= 4^(1005) -= (-1)^1005 -= -1" (mod 5)"$

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 $x=3$

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