The least non-negative remainder, when $5^{61}$ is divided by 7, is

To find the remainder when $5^{61}$ is divided by 7, calculate the first few powers of 5 modulo 7:

$5^1 \equiv 5 \pmod{7}$

$5^2 = 25 \equiv 4 \pmod{7}$

$5^3 = 125 \equiv 6 \pmod{7}$

$5^4 = 625 \equiv 2 \pmod{7}$

$5^5 = 3125 \equiv 3 \pmod{7}$

$5^6 = 15625 \equiv 1 \pmod{7}$

$5^7 = 78125 \equiv 5 \pmod{7}$

The pattern of remainders is: 5, 4, 6, 2, 3, 1, and then repeats with period 6.

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Divide 61 by 6 to determine position in the cycle:

$61 = 6 \times 10 + 1$

The remainder is 1.

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Since $5^6 \equiv 1 \pmod{7}$:

$5^{61} = 5^{6 \times 10 + 1}$

$= (5^6)^{10} \times 5^1$

$\equiv 1^{10} \times 5 \pmod{7}$

$\equiv 5 \pmod{7}$

Therefore, the least non-negative remainder is 5.