Find the remainder when \(7^{131}\) is divided by 100.
Answer: C
We need to find the remainder of \(7^{131} \pmod{100}\). We can find the cycle of remainders for powers of 7.
\(7^1 \equiv 7 \pmod{100}\)
\(7^2 \equiv 49 \pmod{100}\)
\(7^3 = 343 \equiv 43 \pmod{100}\)
\(7^4 = 7^2 \times 7^2 = 49 \times 49 = 2401 \equiv 1 \pmod{100}\)
The cycle of remainders is (7, 49, 43, 1), and the length of the cycle is 4.
To find the remainder for \(7^{131}\), we divide the power (131) by the cycle length (4).
\(131 \div 4 = 32\) with a remainder of 3.
The required remainder is the 3rd term in our cycle, which is 43.