Maximum value of r such that 10!/2*r is an integer

To find the maximum value of r € n such that 10!/2*r is an integer, we need to analyze the prime factorization of the numerator and denominator. Let's break down the problem into smaller steps:

Step 1: Prime factorization of 10!

To find the prime factorization of 10!, we need to multiply all the numbers from 1 to 10 together:

10! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10

Now, let's find the prime factors of each number:

10! = 2^8 * 3^4 * 5^2 * 7

So, the prime factorization of 10! is: 2^8 * 3^4 * 5^2 * 7

Step 2: Prime factorization of 2*r

Since we are looking for the maximum value of r, we need to find the prime factorization of 2*r, where r is a positive integer.

2*r = 2 * r

The prime factorization of 2*r is: 2 * r

Step 3: Canceling common factors

To simplify the expression 10!/2*r, we need to cancel out any common factors between the numerator and the denominator. Since the prime factorization of 10! is known, we can see that 2 is the only common factor between 10! and 2*r.

Now, we can cancel out the common factors:

10!/2*r = (2^8 * 3^4 * 5^2 * 7) / (2 * r)

Cancelling out the common factor of 2, we get:

10!/2*r = 2^7 * 3^4 * 5^2 * 7 / r

Step 4: Finding the maximum value of r

For 10!/2*r to be an integer, the denominator r must be a factor of 2^7 * 3^4 * 5^2 * 7.

To maximize the value of r, we need to take the highest power of 2, 3, 5, and 7 as factors. Therefore, the maximum value of r is:

r = 2^7 * 3^4 * 5^2 * 7

Calculating this value, we get:

r = 128 * 81 * 25 * 7 = 453600

Therefore, the maximum value of r € n such that 10!/2*r is an integer is 453600.