Holiday Contest 2025 - Individual Round - Problem 7


Let a positive integer xx be 1717-ful if there exists a power of 1717 greater than or equal to x\sqrt{x} that divides xx. The number of 1717-ful positive integers less than or equal to 17202517^{2025} is equal to a17ba\cdot 17^b where aa is coprime with 1717. Find a+ba+b.

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Category: Holiday Contest
Points: 5
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