Problem 15 - Entrance Test

For a number to be a perfect square, all the exponents in its prime factorization must be even. First, we need to find the prime factorization of 2023. We can test small prime numbers: 2023 is not divisible by 2, 3 (sum of digits 2+0+2+3=7), 5. Let's try 7: 2023 / 7 = 289. So, 2023 = 7 * 289. Now, we need to factorize 289. We might recognize 289 as 17^2. Indeed, 17 * 17 = 289. So, the prime factorization of 2023 is 7^1 * 17^2. We want 2023n to be a perfect square. Let's substitute the prime factorization: 2023n = (7^1 * 17^2) * n. For 2023n to be a perfect square, all the prime exponents in its factorization must be even. In 7^1 * 17^2: - The exponent of 17 is 2, which is already even. - The exponent of 7 is 1, which is odd. To make the exponent of 7 even, the smallest positive integer n must have at least one factor of 7. If n contains any other prime factors, their exponents must also become even in the product 2023n. To make n the smallest, we only include the necessary prime factors with the minimum required exponents. So, n must contain a factor of 7. If n = 7, then 2023n = 7^1 * 17^2 * 7 = 7^2 * 17^2 = (7 * 17)^2. This is a perfect square (119^2). Any other prime factor in n would make n larger. For example, if n = 7 * k^2 for some integer k > 1, then n would be larger. Therefore, the smallest positive integer n is 7.