Maths Olympiad Prep

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Number theory Difficulty 5.6 AIME, harder Prove it Romania

Find the smallest positive integer which has exactly 20152015 positive divisors.

Solution

The number of positive divisors of an integer NN with the prime factorization p1a1p2a2pnanp_1^{a_1} p_2^{a_2} \cdots p_n^{a_n} is (a1+1)(a2+1)(an+1)(a_1 + 1)(a_2 + 1) \cdots (a_n + 1). Since 2015=12015=5403=13155=3165=513312015 = 1 \cdot 2015 = 5 \cdot 403 = 13 \cdot 155 = 31 \cdot 65 = 5 \cdot 13 \cdot 31, the required number has one of the following forms: a=22014a = 2^{2014}, b=240234b = 2^{402} \cdot 3^4, c=2154312c = 2^{154} \cdot 3^{12}, d=264330d = 2^{64} \cdot 3^{30} or e=23031254e = 2^{30} \cdot 3^{12} \cdot 5^4.

We claim that the smallest number is 230312542^{30} \cdot 3^{12} \cdot 5^4. Indeed, a>ea > e since 21984>312542^{1984} > 3^{12} \cdot 5^4; b>eb > e since 2372>38542^{372} > 3^8 \cdot 5^4; c>ec > e for 2124>542^{124} > 5^4 and d>ed > e for 234318>542^{34} \cdot 3^{18} > 5^4.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.