How many positive integers n are there such that n is a multiple of 5, and the least common multiple of 5! and n equals 5 times the greatest common divisor of 10! and n?

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Let's solve this step by step. We need to find positive integers n that are multiples of 5, where the least common multiple of 5 factorial and n equals 5 times the greatest common divisor of 10 factorial and n. First, we find the prime factorizations. 5 factorial equals 120, which factors as 2 cubed times 3 times 5. 10 factorial equals 3,628,800, which factors as 2 to the 8th times 3 to the 4th times 5 squared times 7. Now we set up our analysis. Let n have prime factorization 2 to the a-2 times 3 to the a-3 times 5 to the a-5 times 7 to the a-7, and so on. Our condition states that the least common multiple of 5 factorial and n equals 5 times the greatest common divisor of 10 factorial and n. We'll analyze this prime by prime using the formulas for LCM and GCD in terms of prime valuations. For each prime p, we get the equation: max of the p-valuation of 5 factorial and the p-valuation of n equals the p-valuation of 5 plus the minimum of the p-valuation of 10 factorial and the p-valuation of n. Let's analyze each prime systematically. For prime 2, we have the equation max of 3 and a-2 equals min of 8 and a-2. By case analysis: if a-2 is less than 3, we get 3 equals a-2, which is impossible. If a-2 is greater than 8, we get a-2 equals 8, also impossible. The equation only holds when 3 is less than or equal to a-2 which is less than or equal to 8, giving us 6 possible values. Similarly, for prime 3, we get max of 1 and a-3 equals min of 4 and a-3, which holds when 1 is less than or equal to a-3 which is less than or equal to 4, giving us 4 possible values. Now for the critical cases. For prime 5, we have max of 1 and a-5 equals 1 plus min of 2 and a-5. Since n must be a multiple of 5, we need a-5 at least 1. Testing a-5 equals 1 gives 1 equals 2, which is false. Testing a-5 equals 2 gives 2 equals 3, also false. For a-5 greater than 2, we get a-5 equals 3, which works. So a-5 must equal 3, giving us exactly 1 choice. For prime 7, we have max of 0 and a-7 equals min of 1 and a-7, which holds when a-7 equals 0 or 1, giving us 2 choices. For primes greater than 7, we have max of 0 and a-p equals min of 0 and a-p, which only holds when a-p equals 0. This means n cannot have any prime factors greater than 7. Therefore, n must have the form 2 to the a-2 times 3 to the a-3 times 5 to the a-5 times 7 to the a-7. Summarizing our constraints: a-2 has 6 possible values from 3 to 8, a-3 has 4 possible values from 1 to 4, a-5 has exactly 1 value which is 3, and a-7 has 2 possible values 0 or 1. The total number of solutions is 6 times 4 times 1 times 2, which equals 48.