The non-negative remainder when $7^{30}$ is divided by 5 is

When dividing 7 by 5, the remainder is 2.

$7 = 5 \times 1 + 2$

So when finding remainders, 7 can be replaced with 2:

$7^{30} \equiv 2^{30} \pmod{5}$

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Powers of 2 divided by 5 give the following remainders:

$2^1 \equiv 2 \pmod{5}$

$2^2 \equiv 4 \pmod{5}$

$2^3 \equiv 3 \pmod{5}$

$2^4 \equiv 1 \pmod{5}$

$2^5 \equiv 2 \pmod{5}$

The remainders follow a cycle of length 4: $\;2 \to 4 \to 3 \to 1 \to 2 \to 4 \to \ldots$

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Since the cycle repeats every 4 powers, divide 30 by 4:

$30 = 4 \times 7 + 2$

A remainder of 2 means $2^{30}$ falls on the same position as $2^2$ in the cycle.

$2^{30}$

$= (2^4)^7 \times 2^2$

$\equiv 1^7 \times 4 \pmod{5}$

$\equiv 4 \pmod{5}$

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$\therefore \; 7^{30} \equiv 2^{30} \equiv 4 \pmod{5}$

The non-negative remainder when $7^{30}$ is divided by 5 is $4$.