Understanding the Problem: Remainder Calculation

The question asks us to find the remainder when the number \(3^{10}\) is divided by 7. This is a problem in modular arithmetic, often expressed as finding the value of \(3^{10} \pmod{7}\).

Method 1: Finding a Pattern in Remainders

We can calculate the remainders of the first few powers of 3 when divided by 7 to see if there's a repeating pattern:

• \(3^1 \div 7\) has a remainder of 3. So, \(3^1 \equiv 3 \pmod{7}\).
• \(3^2 = 9\). \(9 \div 7\) has a remainder of 2. So, \(3^2 \equiv 2 \pmod{7}\).
• \(3^3 = 27\). \(27 \div 7\) has a remainder of 6 (\(27 = 3 \times 7 + 6\)). So, \(3^3 \equiv 6 \pmod{7}\).
• \(3^4 = 81\). \(81 \div 7\) has a remainder of 4 (\(81 = 11 \times 7 + 4\)). So, \(3^4 \equiv 4 \pmod{7}\).
• \(3^5 = 243\). \(243 \div 7\) has a remainder of 5 (\(243 = 34 \times 7 + 5\)). So, \(3^5 \equiv 5 \pmod{7}\).
• \(3^6 = 729\). \(729 \div 7\) has a remainder of 1 (\(729 = 104 \times 7 + 1\)). So, \(3^6 \equiv 1 \pmod{7}\).

Notice that once we reach a remainder of 1 (\(3^6 \equiv 1 \pmod{7}\)), the pattern of remainders will start repeating. The cycle of remainders is (3, 2, 6, 4, 5, 1), and the length of this cycle is 6.

Applying the Pattern to \(3^{10}\)

To find the remainder of \(3^{10}\) divided by 7, we need to figure out where in the cycle the 10th power falls. We do this by dividing the exponent (10) by the cycle length (6):

\(10 \div 6 = 1 \text{ \textbf{with a remainder of}} 4\)

This remainder of 4 tells us that \(3^{10}\) will have the same remainder as the 4th number in our cycle of remainders.

The 4th remainder in the cycle (3, 2, 6, 4, 5, 1) is 4.

Therefore, the remainder when \(3^{10}\) is divided by 7 is 4.

Method 2: Using Fermat's Little Theorem

Fermat's Little Theorem is a useful tool in modular arithmetic. It states that if '\(p\)' is a prime number, then for any integer '\(a\)' not divisible by '\(p\)', we have:

\(a^{p-1} \equiv 1 \pmod{p}\)

In our case, \(a=3\) and \(p=7\). Since 7 is prime and 3 is not divisible by 7, we can apply the theorem:

\(3^{7-1} \equiv 3^6 \equiv 1 \pmod{7}\)

Now we want to find \(3^{10} \pmod{7}\). We can rewrite the exponent 10 using the exponent from the theorem (6):

\(10 = 6 + 4\)

So, we can write \(3^{10}\) as:

\(3^{10} = 3^{6+4} = 3^6 \times 3^4\)

Now, let's find the remainder modulo 7:

\(3^{10} \equiv (3^6 \times 3^4) \pmod{7}\)

Since we know \(3^6 \equiv 1 \pmod{7}\), we substitute this in:

\(3^{10} \equiv (1 \times 3^4) \pmod{7}\)

\(3^{10} \equiv 3^4 \pmod{7}\)

Now, we just need to calculate \(3^4\) and find its remainder when divided by 7:

\(3^4 = 81\)

\(81 \div 7 = 11 \text{ \textbf{with a remainder of}} 4\)

So, \(3^4 \equiv 4 \pmod{7}\).

Putting it all together:

\(3^{10} \equiv 4 \pmod{7}\)

Both methods show that the remainder when \(3^{10}\) is divided by 7 is 4.