Solution
The unit digits of powers of 3 repeat in the block 3, 9, 7, 1 with period 4, and $$2025 = 4 \times 506 + 1$$, so $$3^{2025}$$ ends in 3. The unit digits of powers of 7 repeat as 7, 9, 3, 1, and 2024 is a multiple of 4, so $$7^{2024}$$ ends in 1. The product therefore ends in $$3 \times 1 = 3$$.
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