Try each problem before reading the solutions. Write down the step that makes your argument complete.
Problems
1. Find the last digit of 7²⁰²⁶.
2. Find the smallest positive integer that leaves remainder 2 upon division by 5 and remainder 3 upon division by 7.
3. Prove that no perfect square leaves remainder 2 when divided by 4.
Solutions
1. The last digits of successive powers of 7 cycle through 7, 9, 3, 1. Since 2026 leaves remainder 2 when divided by 4, the last digit is 9.
2. The first positive candidates that leave remainder 3 modulo 7 are 3, 10 and 17. The first two fail the modulo 5 condition, and 17 leaves remainder 2, so 17 is minimal.
3. An even square (2k)² is divisible by 4. An odd square (2k + 1)² equals 4k(k + 1) + 1. The only possible remainders are 0 and 1.
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