Understanding the Remainder Calculation

The question asks us to find the remainder when the number \(2^{120}\) is divided by \(7\). This is a problem that can be solved using the concepts of modular arithmetic, specifically by looking for patterns in the remainders of powers of \(2\) when divided by \(7\). We want to find the value of \(2^{120} \pmod{7}\).

Exploring Powers of 2 Modulo 7

Let's calculate the first few powers of \(2\) and find their remainders when divided by \(7\):

• \(2^1 = 2\). The remainder when \(2\) is divided by \(7\) is \(2\). So, \(2^1 \equiv 2 \pmod{7}\).
• \(2^2 = 4\). The remainder when \(4\) is divided by \(7\) is \(4\). So, \(2^2 \equiv 4 \pmod{7}\).
• \(2^3 = 8\). The remainder when \(8\) is divided by \(7\) is \(1\). So, \(2^3 \equiv 1 \pmod{7}\).

We've found a pattern! Since \(2^3\) gives a remainder of \(1\), the remainders will start repeating from this point.

Let's check the next power:

• \(2^4 = 2^3 \times 2^1\). Using modular arithmetic, \(2^4 \equiv 1 \times 2 \equiv 2 \pmod{7}\).
• \(2^5 = 2^3 \times 2^2\). Using modular arithmetic, \(2^5 \equiv 1 \times 4 \equiv 4 \pmod{7}\).
• \(2^6 = 2^3 \times 2^3\). Using modular arithmetic, \(2^6 \equiv 1 \times 1 \equiv 1 \pmod{7}\).

The pattern of the remainders is \(2, 4, 1\), and this cycle repeats every \(3\) powers.

Applying the Cyclical Pattern

To find the remainder of \(2^{120}\) divided by \(7\), we need to know where in this cycle the \(120\)-th power falls. We can determine this by looking at the exponent, \(120\), and finding its remainder when divided by the cycle length, which is \(3\).

Let's divide the exponent \(120\) by the cycle length \(3\):

\(\frac{120}{3} = 40\)

The remainder is \(0\). This means that \(120\) is a multiple of \(3\). When the exponent is a multiple of the cycle length, the remainder matches the last element in the cycle pattern.

In our pattern \((2, 4, 1)\), the third element corresponds to \(2^3\), which has a remainder of \(1\). Alternatively, since \(120 \equiv 0 \pmod{3}\), we can write:

\(2^{120} = 2^{3 \times 40} = (2^3)^{40}\)

Now, apply the modulo \(7\) operation:

\(2^{120} \equiv (2^3)^{40} \pmod{7}\)

Since we know \(2^3 \equiv 1 \pmod{7}\), substitute this value:

\(2^{120} \equiv (1)^{40} \pmod{7}\)

Any power of \(1\) is still \(1\):

\(1^{40} = 1\)

Therefore:

\(2^{120} \equiv 1 \pmod{7}\)

Conclusion

The remainder when \(2^{120}\) is divided by \(7\) is \(1\). This corresponds to the first option.