Discussion
What is the remainder of \(5555^{2222} + 2222^{5555}\) divided by 7?
Answer: A
We find the remainder of each part separately modulo 7.
Part 1: \(5555^{2222} \pmod{7}\). First, simplify the base: \(5555 \div 7\). \(5555 = 793 \times 7 + 4\). So, \(5555 \equiv 4 \pmod{7}\). The problem becomes \(4^{2222} \pmod{7}\). The cycle of powers of 4 mod 7 is \(4^1=4, 4^2=16\equiv 2, 4^3=8\equiv 1\). The cycle length is 3. We find the remainder of the power \(2222 \div 3\). Sum of digits is 8, so remainder is 2. The required remainder is the 2nd in the cycle (4, 2, 1), which is 2.
Part 2: \(2222^{5555} \pmod{7}\). Simplify the base: \(2222 \div 7\). \(2222 = 317 \times 7 + 3\). So, \(2222 \equiv 3 \pmod{7}\). The problem is \(3^{5555} \pmod{7}\). Cycle for 3: \(3^1=3, 3^2=9\equiv 2, 3^3\equiv 6, 3^4\equiv 18\equiv 4, 3^5\equiv 12\equiv 5, 3^6\equiv 15\equiv 1\). Cycle length is 6. Remainder of power \(5555 \div 6\). \(5555 = 925 \times 6 + 5\). Remainder is 5. The required remainder is the 5th in the cycle, which is 5.
Total remainder = \((2 + 5) \pmod{7} = 7 \pmod{7} = 0\).
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