If n is a positive integer, n! is the product of the first n positive integers. For example, 4! = 4 x 3 x 2 x 1 =24. If u and v are positive integers and u!=v! x 53, then v could equal
A. 6
B. 8
C. 56
D. 57
53 is a prime, so it cannot appear in any factorial less than 53!
I don't see how v can be anything other than 52, but that's not one of the choices.
Hmmmm...
To solve this problem, we need to find a value for v that satisfies the given equation u! = v! x 53.
First, let's rewrite the equation using the factorial definition: u x (u-1) x (u-2) x ... x 3 x 2 x 1 = v x (v-1) x (v-2) x ... x 3 x 2 x 1 x 53.
We can see that 53 is a prime number and it only appears on the right side of the equation. To have a solution, we need to find a value for v where the left side of the equation doesn't have any prime factor equal to 53.
Looking at the answer choices, we can check each one to see if it satisfies this condition:
A. If v = 6, then 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720. 720 is not divisible by 53, so this is a possible value for v.
B. If v = 8, then 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320. 40320 is not divisible by 53, so this is a possible value for v.
C. If v = 56, then 56! = 56 x 55 x ... x 3 x 2 x 1. In this case, 56! is divisible by 53 because 56 = 53 + 3 (notice that 53 is a prime factor of 56!). Therefore, this value of v doesn't satisfy the condition.
D. If v = 57, then 57! = 57 x 56 x ... x 3 x 2 x 1. Here, 57! is divisible by 53 because 57 = 53 + 4 (again, notice that 53 is a prime factor of 57!). So, this value of v also doesn't satisfy the condition.
Hence, the possible values for v are A. 6 and B. 8.