MOPSS, 19th July 2025

For positive integers \(x, y\), note that \[ \frac{1}{x} + \frac{1}{y} = \frac{1}{n} \] holds if and only if \[ (x - n)(y - n) = n^2 \] holds. Observe that if \(x, y\) are positive integers satisfying the given equation, then \(x > n\) and \(y > n\) holds. This shows that the solutions of the given equation over the positive integers are in one-to-one correspondence with the pairs of positive integers \((a, b)\) such that \(ab = n^2\), through the map \[ (a + n, b + n) \leftrightarrow (a, b). \] Hence, the set of positive integers \(n\) satisfying \(S(n) = 5\) is equal to the set of positive integers \(n\) such that \(n^2\) has precisely \(5\) positive divisors. Note that any such integer \(n\) is larger than \(1\). Writing \(n\) as a product of powers of distinct primes, it follows that \(n^2\) has precisely \(5\) positive divisors if and only if \(n\) is the square of a prime. Indeed, if \(p_1, \dots, p_r\) are distinct primes, and \(a_1, \dots, a_r\) are positive integers, then the integer \((p_1^{a_1} \dots p_r^{a_r})^2\) has precisely \(5\) positive divisors if and only if \[ (2a_1 + 1)(2a_2 + 1) \dots (2a_r + 1) = 5 \] holds, which is equivalent to \(r = 1, a_1 = 2\). This proves that the positive integers satisfying \(S(n) = 5\) are precisely the squares of the primes.

Note that if \(n \geq 1\), and \(a_1, \dots, a_n\) are distinct and pairwise coprime positive integers such that \(a_i \geq 2\) for all \(i\), then one of them admits a prime factor which is at least as large as the \(n\)-th prime. Indeed, for each \(i\), if we fix a prime divisor \(p_i\) of \(a_i\), then using that \(a_1, \dots, a_n\) are pairwise coprime, it follows that \(p_1, \dots, p_n\) are distinct primes, and hence the largest among them is at least as large as the \(n\)-th prime. Consequently, if \(n \geq 2\), and \(a_1, \dots, a_n\) are distinct and pairwise coprime positive integers, then at least \((n - 1)\) of them are greater than \(1\), and hence one of them is divisible by a prime at least as large as the \((n-1)\)-st prime. If possible, let us assume that the elements of \(A\) are pairwise coprime. Hence, \(A\) contains an element \(x\) which is divisible by a prime at least as large as the \(15\)th prime. Similarly, the set \(A\setminus \{x\}\) contains an element \(y\) which is divisible by a prime at least as large as the \(14\)th prime. Since the \(14\)th and \(15\)th primes are \(43, 47\) respectively, it follows that \(x\geq 47, y \geq 43\), and consequently, \[ xy \geq 47 \cdot 43 = 2021 >1994, \] which contradicts the hypothesis. This proves that there are two numbers \(a\) and \(b\) in \(A\) which are not relatively prime.

Since \(a_1a_2, a_1a_2a_3a_4, a_1a_2a_3a_4a_5a_6\) are divisible by \(2\), it follows that \(a_2, a_4, a_6\) are even, and hence they are equal to \(2, 4, 6\) in some order. Using that \(a_1a_2a_3a_4a_5\) is divisible by \(5\), we get that \(a_5 = 5\). So \(a_1, a_3\) are equal to \(1, 3\) in some order. Using that \(a_1a_2a_3\) is a multiple of \(3\), we obtain \[ a_1+ a_2+ a_3\equiv 0 \bmod 3, \] which yields \[ a_2 \equiv 2 \bmod 3, \] and hence, \(a_2 = 2\) holds. Note that \(1234, 3214\) are not divisible by \(4\). This shows that \(a_4 = 6\), and hence \(a_6 = 4\). It follows that \(a_1a_2a_3a_4a_5a_6\) is equal to \(321654\), or \(123654\). Note that the integers \(321654, 123654\) satisfy the required conditions too. This proves that \(321654, 123654\) are precisely all the \(6\)-digit numbers satisfying the given condition.

Note that among \(18\) consecutive three digit numbers, there is an integer divisible by \(18\). Denote it by \(n = 100a + 10b + c\) with \(a, b, c\) denoting integers lying between \(0\) and \(9\). It follows that \(9\) divides \(n\), and hence \(9\) divides \(a + b + c\). This shows that \(a+b+c\) is equal to one of \(9, 18, 27\). Note that \(a + b + c = 27\) holds only if \(n = 999\). Since \(18\) divides \(n\), it follows that \(a + b + c \neq 27\), and hence, \(a + b + c\) is equal to one of \(9, 18\). This proves that \(a + b + c\) divides \(n\).

Note that \(p_1p_2 \dots p_n + 1\) is odd for any \(n\geq 1\), and hence \(p_n\) is odd for any \(n\geq 2\). Since \(p_1 = 2\) and \(p_2 = 3\), it follows that for any \(n\geq 2\), the integer \(p_1p_2 \dots p_n + 1\) is not divisible by any one of \(2\) and \(3\). So the least prime divisor of \(p_1p_2 \dots p_n + 1\) is at least \(5\) for any \(n\geq 2\). If possible, suppose \(5\) is the largest prime divisor of \(p_1p_2 \dots p_n + 1\) for some integer \(n\geq 2\). This yields \[ p_1p_2 \dots p_n + 1 = 5^k \] for some \(k\geq 1\). This implies that \(4\) divides \(p_1p_2 \cdots p_n\), which is impossible since \(p_1 = 2\), and \(p_r\) is odd for any integer \(r \geq 2\). This shows that \(p_{n+1}\) is not equal to \(5\) for any integer \(n\geq 2\). Consequently, it follows that \(p_n\neq 5\) for any integer \(n\geq 1\).

Let \(x, y\) be integers such that \(17\) divides both the expressions \(x^2 - 2xy + y^2 - 5x + 7y\) and \(x^2 - 3xy + 2y^2 + x - y\). Note that \[ x^2 - 3xy + 2y^2 + x - y = (x-y)( x-2y+1), \] which is divisible by \(17\). It follows that \[ x \equiv y \bmod 17, \quad \text{ or } x \equiv 2y -1 \bmod 17 \] holds. Let us consider the case that \(x \equiv y \bmod 17\). It follows that \[ x^2 - 2xy + y^2 - 5x + 7y \equiv (x-y)^2 - 5x+ 7y \equiv 2y \bmod 17. \] Since \(17\) divides \(x^2 - 2xy + y^2 - 5x + 7y\), we get \(2y \equiv 0 \bmod 17\), which yields \(x\equiv y \equiv 0 \bmod 17\), and hence \(17\) divides \(xy - 12x + 15y\). Let us consider the case that \(x \equiv 2y -1 \bmod 17\). Using \(x^2 - 2xy + y^2 - 5x + 7y\equiv 0 \bmod 17\), we obtain \[ (2y -1 )^2 - 2(2y -1 )y + y^2 - 5(2y-1) + 7y\equiv 0 \bmod 17, \] which yields \(y^2-5y + 6 \equiv 0 \bmod 7\). This implies that \((y-2)(y-3)\equiv 0 \bmod 17\). This shows that either \(x\equiv 3\bmod 17, y \equiv 2 \bmod 17\) holds, or \(x \equiv 5 \bmod 17, y \equiv 3 \bmod 17\) holds. If \(x\equiv 3\bmod 17, y \equiv 2 \bmod 17\) holds, then \[ xy - 12x + 15y \equiv 6 - 36 + 30 \equiv 0 \bmod 17 \] holds. If \(x \equiv 5 \bmod 17, y \equiv 3 \bmod 17\) holds, then we obtain \[ xy - 12x + 15y \equiv 15 - 60 + 45 \equiv 0 \bmod 17. \] This proves that \(17\) divides \(xy - 12x + 15y\).