# How do you find the remainder of 3^983 divided by 5?

A remainder of

You can observe a pattern that repeats over four numbers if you look at the final digit.

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There's a pattern in their final numbers:

Every fourth power, it repeats.

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To find the remainder of (3^{983}) divided by 5, we can use the concept of modular arithmetic and Euler's theorem. Euler's theorem states that if (a) and (n) are coprime integers (i.e., they have no common factors other than 1), then (a^{\phi(n)} \equiv 1 \pmod{n}), where (\phi(n)) is Euler's totient function, which gives the number of positive integers less than (n) that are coprime to (n).

In this case, since 5 is prime, (\phi(5) = 5 - 1 = 4). So, according to Euler's theorem: [3^{4} \equiv 1 \pmod{5}]

Now, we need to find the remainder of (3^{983}) divided by 4: [983 \equiv 3 \pmod{4}]

So, (3^{983} \equiv 3^3 \pmod{5}): [3^{983} \equiv 27 \pmod{5}] [3^{983} \equiv 2 \pmod{5}]

Therefore, the remainder of (3^{983}) divided by 5 is 2.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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