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\phi (n)= 2^5 / 47 n , prime factorization

Source: China Northern MO 2015 grade 10 p3 CNMO

May 5, 2024
number theoryprime factorization

Problem Statement

If n=p1a1,p2a2...psasn=p_1^{a_1},p_2^{a_2}...p_s^{a_s} then ϕ(n)=n(1−1p1)(1−1p2)...(1−1ps)\phi (n)=n \left(1- \frac{1}{p_1}\right)\left(1 - \frac{1}{p_2}\right)...\left(1- \frac{1}{p_s}\right). Find the smallest positive integer nn such that ϕ(n)=2547n.\phi (n)=\frac{2^5}{47}n.
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