Step-by-step Solution:

Quick Solution: Finding the Remainder

The problem is to find the remainder when $2312^{31}$ is divided by 5.

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1. Simplify the Base

First, we can simplify the problem by finding the remainder of the base (2312) when divided by 5. A number's remainder when divided by 5 is the same as its last digit's remainder.

$2312 \equiv 2 \pmod{5}$

Therefore, the original problem $2312^{31} \pmod{5}$ is equivalent to solving $2^{31} \pmod{5}$.

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Method 1: Finding the Cyclical Pattern

2. Find the Cycle of Remainders

Let's look at the pattern of the remainders for the first few powers of 2 when divided by 5:

• $2^1 = 2 \equiv 2 \pmod{5}$
• $2^2 = 4 \equiv 4 \pmod{5}$
• $2^3 = 8 \equiv 3 \pmod{5}$
• $2^4 = 16 \equiv 1 \pmod{5}$

The pattern of remainders repeats in a cycle of length 4: (2, 4, 3, 1).

3. Find the Position in the Cycle

To find which remainder corresponds to the exponent 31, we find the remainder of 31 when divided by the cycle length (4).

$31 \div 4 = 7$ with a remainder of 3.

This means we need the 3rd number in our cycle. The 3rd number in the cycle (2, 4, 3, 1) is 3.

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Method 2: Using Euler's Theorem (Faster)

Euler's totient theorem is a more advanced method that is very fast for large exponents. It states that if 'a' and 'n' are coprime (have no common factors other than 1), then $a^{\phi(n)} \equiv 1 \pmod{n}$.

1. Find $\phi(n)$: Here, n=5, which is a prime number. For any prime number p, $\phi(p) = p-1$. So, $\phi(5) = 4$.
   
2. Apply the Theorem: Since 2 and 5 are coprime, we know that $2^{\phi(5)} \equiv 1 \pmod{5}$, which means $2^4 \equiv 1 \pmod{5}$. This mathematically confirms our cycle length of 4.
   
3. Reduce the Exponent: We rewrite the exponent 31 in terms of 4. We know that $31 = 4 \times 7 + 3$.
   
4. Solve:
   

   $2^{31} = 2^{(4 \times 7 + 3)} = (2^4)^7 \times 2^3$
   

   Now we take the remainder of the whole expression:
   

   $(2^4)^7 \times 2^3 \equiv (1)^7 \times 8 \pmod{5}$
   
   $\equiv 1 \times 3 \pmod{5}$
   
   $\equiv 3 \pmod{5}$
   

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Conclusion

Both methods show that the remainder is 3.

Final Answer: The remainder when $2312^{31}$ is divided by 5 is 3.