A couple of days ago, the following question was posted on the MathHelp subreddit.
6 is divisible by all integers up to its half: 1, 2, and 3. Are there any numbers >6 with this property?
In other words, cases where x is divisible by all integers from 1 to x/2.
Something tells me 6 is the highest but I have no idea how to go about proving it.
I gave the following response.
Let me rephrase your question , as I understand it, more precisely. Does there exist a natural numbers such that for all and for all , .? Did I get that correct?
and suggestion
BTW, it should be straightforward to prove that no such natural number exists, by considering the prime decomposition of .
In this post I will give a precise statement of the result and a full proof of it.
Proof. It is easy to verify by inspection that , and have this property while and do not. It is shown that and are the only numbers greater than or equal to with this property by consideration of the prime decomposition of
Assume that has the property. For , . Hence and the prime decomposition of takes the form and . It is assumed without loss of generality that . Clearly, Observe that . Thus by assumption . It follows that the prime decomposition of takes the form where . Substitution from and followed by a factorization gives If the product of two natural numbers equals , then each of the numbers equals . It follows that and From equation , one obtains . Thus equation can be written as and equation becomes If , then gives and . In this case On the other hand, suppose that . Then consideration of gives . However, , consequently . Again, consideration of yields Therefore and . Thus in this case This concludes the proof. ◻