How many ordered pairs of positive integers 1≤k≤n≤50 are there, such that k divides n, and
(n/k)!=n!/k!?
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To understand the problem, let's break it down step by step.
We need to find the number of ordered pairs of positive integers (k, n) where 1 ≤ k ≤ n ≤ 50, such that k divides n, and (n/k)! = n!/k!.
Let's first simplify the given condition: (n/k)! = n!/k!.
By expanding the factorials, we get:
(n/k)(n/k - 1)(n/k - 2)...(1) = (n)(n - 1)(n - 2)...(1) / (k)(k - 1)(k - 2)...(1)
Simplifying this further, we have:
(n)(n - k)(n - 2k)(n - 3k)...(1) = (n)(n - 1)(n - 2)...(1) / (k)(k - 1)(k - 2)...(1)
Next, we can see that both the numerator and denominator of the right side of the equation contain the same terms up to k - 1, which cancels out:
(n)(n - k)(n - 2k)(n - 3k)...(k + 1) = (n)(n - 1)(n - 2)...(k + 1)
To find the number of solutions, it is helpful to consider the prime factorization of n and k.
Let p be a prime factor of k, and let p^a be the highest power of p that divides k. Then, p^(a+1) must divide n.
Based on this observation, for each prime factor p of k, we need to choose the highest exponent a such that p^(a+1) is less than or equal to 50.
For example, let's consider the prime factor 2. The highest power of 2 that divides k can be 2^0, 2^1, 2^2, ..., 2^5. Therefore, we have 6 options for the exponent of 2.
We repeat this process for all prime factors of k, and multiply the number of options for each prime factor together to get the total number of solutions.
Let's go step by step to calculate the total number of solutions:
1. Calculate the prime factorization of k and determine the highest exponent a for each prime factor that is less than or equal to 50.
2. For each exponent a, calculate the number of options (a + 1).
3. Multiply all the options together to get the total number of solutions.
As the prime factorization and calculation can be time-consuming, it is recommended to write a program or use a mathematical software to perform these calculations.