Factorial of Smallest Multiple Divisible by Square!
Find the smallest positive integer value of x such that for every positive integer a greater than 1, ( x a ) ! is divisible by a 2 .
If there is no such positive integer value x , then enter 0.
The answer is 2.
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2 solutions
The problem can be generalised as ( x a ) ! is divisible by a k if x = k
Can you add a proof of the generalization?
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This generalization is instantly obvious from the fact that:
n ! ( r ! ) k ( n r ) ! is an integer for n , r ≥ 1 .
This fact can be proved by induction on n .
By the way, k isn't necessarily the minimum positive integer value of x such that ( x a ) ! is divisible by a k .
Yes. But k isn't necessarily the minimum value of x so that ( x a ) ! is divisible by a k .
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If k is prime then ??
The answer is 2 .
First , let's see that it works. For any positive integer a , we have ( 2 a ) ! = 1 × 2 × 3 × … … … × a × … … … × ( 2 a − 1 ) × ( 2 a ) , where the factors a and 2 a confirms the divisibility of ( 2 a ) ! by a 2 .
Second , the only smaller candidate is 1 . Now showing " 1 doesn't work" will suffice. Just think of a prime value of a .