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There are 2025 cards numbered from 01 to 2025 (i.e. 1, 2, 3, ..., 2025). One card is drawn at random, then the probability that the number on the selected card leaves remainder of 25 when it divides 2025, is
If n leaves remainder 25 on dividing 2025, then n divides $$2025 - 25 = 2000$$ and $$n > 25$$. Since $$2000 = 2^4 \times 5^3$$ has $$5 \times 4 = 20$$ divisors, and 9 of them (1, 2, 4, 5, 8, 10, 16, 20, 25) are at most 25, there are 11 valid numbers. The probability is $$\frac{11}{2025}$$.
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