Question

Find the remainder when \(2^{88} \times 5^{41}\) is divided by 7.

Hint:

When finding remainders of large powers \(a^b \pmod{p}\), find the cycle length \(k\) such that \(a^k \equiv 1 \pmod{p}\). Then, the problem reduces to calculating \(a^{b \pmod{k}} \pmod{p}\). Fermat's Little Theorem is a powerful tool for finding this cycle length when \(p\) is prime.

Answer 1

Step 1: Understanding the Concept:
This problem requires finding the remainder of a large exponential expression, which is a typical modular arithmetic problem. We can use the properties of modular arithmetic and Fermat's Little Theorem to simplify the calculation. Fermat's Little Theorem states that if \(p\) is a prime number, then for any integer \(a\) not divisible by \(p\), we have \(a^{p-1} \equiv 1 \pmod{p}\).
Step 2: Detailed Explanation:
We need to compute \( (2^{88} \times 5^{41}) \pmod{7} \). We can compute the remainders for each part separately.
For \(2^{88} \pmod{7}\):
Let's find the cycle of remainders of powers of 2 when divided by 7. \[ 2^1 \equiv 2 \pmod{7} \] \[ 2^2 \equiv 4 \pmod{7} \] \[ 2^3 \equiv 8 \equiv 1 \pmod{7} \] The cycle length is 3. To find the remainder of \(2^{88}\), we need to find the remainder of the exponent 88 when divided by the cycle length 3. \[ 88 \div 3 = 29 \text{ with a remainder of } 1 \] So, \(88 \equiv 1 \pmod{3}\). Therefore, \(2^{88} \equiv 2^1 \equiv 2 \pmod{7}\).
For \(5^{41} \pmod{7}\):
By Fermat's Little Theorem, since 7 is prime, \(5^{7-1} \equiv 5^6 \equiv 1 \pmod{7}\). The cycle length is 6. To find the remainder of \(5^{41}\), we need to find the remainder of the exponent 41 when divided by 6. \[ 41 \div 6 = 6 \text{ with a remainder of } 5 \] So, \(41 \equiv 5 \pmod{6}\). Therefore, \(5^{41} \equiv 5^5 \pmod{7}\). Let's calculate \(5^5 \pmod{7}\): \[ 5^1 \equiv 5 \pmod{7} \] \[ 5^2 \equiv 25 \equiv 4 \pmod{7} \] \[ 5^3 \equiv 5 \times 4 = 20 \equiv 6 \pmod{7} \] \[ 5^4 \equiv 5 \times 6 = 30 \equiv 2 \pmod{7} \] \[ 5^5 \equiv 5 \times 2 = 10 \equiv 3 \pmod{7} \] So, \(5^{41} \equiv 3 \pmod{7}\).
Step 3: Final Answer:
Now, we combine the results. \[ (2^{88} \times 5^{41}) \pmod{7} \equiv (2 \times 3) \pmod{7} \] \[ \equiv 6 \pmod{7} \] The remainder is 6.