Maths Olympiad Prep

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, 2022

Number theory Difficulty 5.0 AIME Prove it Bulgaria

A positive integer nn having 2022 divisors 1=d1<d2<<d2022=n1 = d_1 < d_2 < \dots < d_{2022} = n is called nice if 3d2017+2d2019=n3d_{2017} + 2d_{2019} = n. Find all nice integers.

Solution

It follows from diid_i \ge i and d6d2017=d4d2019=nd_6 d_{2017} = d_4 d_{2019} = n that
2n=6d2017+4d2019d6d2017+d4d2019=2n. 2n = 6 d_{2017} + 4 d_{2019} \le d_6 d_{2017} + d_4 d_{2019} = 2n.
Therefore d6=6d_6 = 6 implying di=id_i = i for i=1,2,3,4,5,6i = 1, 2, 3, 4, 5, 6 and thus nn is divisible by 22352^2 \cdot 3 \cdot 5. Since 2022=233372022 = 2 \cdot 3 \cdot 337 and 337337 is a prime number it follows that n=2x3y5zn = 2^x \cdot 3^y \cdot 5^z for x2x \ge 2. All nice numbers are:
2235336,2233365,2336352,2336325. 2^2 \cdot 3 \cdot 5^{336},\quad 2^2 \cdot 3^{336} \cdot 5,\quad 2^{336} \cdot 3 \cdot 5^2,\quad 2^{336} \cdot 3^2 \cdot 5.

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