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 find the number of ordered pairs of positive integers that satisfy the given conditions, we need to break down the problem into smaller steps. Let's proceed step by step:
Step 1: Understand the conditions
- We need to find ordered pairs of positive integers (k, n) such that k divides n, meaning n is divisible by k without any remainder.
- Additionally, we need to check if (n/k)! is equal to n!/k!.
Step 2: Simplify the conditions
- Since k divides n, we can express n as a multiple of k: n = a * k, where a is a positive integer.
- Using this expression, let's rephrase the condition (n/k)! = n!/k!:
- (n/k)! = ((a * k) / k)! = a!,
- n!/k! = (a * k)! / k! = a! * (a * k) * (a * k - 1) * ... * (a * k - k + 1).
- Now we have: a! = a! * (a * k) * (a * k - 1) * ... * (a * k - k + 1).
Step 3: Analyze the equation
- To solve the last equation, we have two possibilities:
- Case 1: a = 1, which leads to k = 1.
- Case 2: a > 1, in which we can cancel out the common factors from both sides. This leads to k = 1, which is not a possible case according to the problem conditions.
Step 4: Analyze k = 1
- In the case where k = 1, the value of n does not matter, as any positive integer n will be divisible by 1 without any remainder.
- Therefore, for every positive integer n within the given range of 1≤k≤n≤50, we have a valid ordered pair (k, n), where k = 1.
Step 5: Count the number of ordered pairs
- Since k = 1 for all valid solutions, the number of ordered pairs is equal to the number of positive integers n within the given range of 1≤n≤50.
- Counting the integers from 1 to 50, we find that there are 50 such integers.
- Hence, there are a total of 50 ordered pairs of positive integers (k, n) that satisfy the given conditions.
Therefore, the answer is 50.