The least non-negative remainder when $3^{128}$ is divided by 7 is:

To find the remainder when $3^{128}$ is divided by 7, calculate the first few powers of 3 modulo 7:

$3^1 = 3$ → Remainder = $3$

$3^2 = 9$ → Remainder = $2$

$3^3 = 27$ → Remainder = $6$

$3^4 = 81$ → Remainder = $4$

$3^5 = 243$ → Remainder = $5$

$3^6 = 729$ → Remainder = $1$

When $3^6$ gives remainder 1, the pattern repeats every 6 powers.

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To determine where 128 falls in this cycle, divide 128 by 6:

$128 \div 6 = 21$ remainder $2$

Therefore:

$128 = 6 \times 21 + 2$

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Express $3^{128}$ using this decomposition:

$3^{128} = 3^{6 \times 21 + 2}$

$= 3^{6 \times 21} \times 3^2$

$= (3^6)^{21} \times 3^2$

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Since $3^6$ gives remainder 1 when divided by 7:

$(3^6)^{21}$ gives remainder $1^{21} = 1$

$3^2 = 9$ gives remainder $2$

Therefore, the remainder of $3^{128}$ when divided by 7:

$1 \times 2 = 2$

The least non-negative remainder when $3^{128}$ is divided by 7 is $2$.