Find all integers \(n\) such that the arithmetic mean of all its factors is an integer.
Thanks so much!
Every odd prime number has two factors and only two factors, 1 and the prime number.
The result of adding 1 to the odd prime number is an even number; dividing this
answer by 2 results in an integer.
Now, you have an infinite number of answers.
And yet, there are more; some even numbers also fulfill the requirement ...
Oh! That's really cool! I haven't thought of the odd primes.
I also assume that some non-odd primes work and as you say, some even numbers work too. Is there a specific way of describing them or is it just at random?
Thank you so much,geno3141!
I think your answer is cool too :)
Thank you for responsing to Gino's answer so nicely.
You need to think.
Your statement: "I also assume that some non-odd primes work"
1) Firstly, tell me some non-odd primes.
2) second show me with one of them what the average of its factors is.
3) discuss your answers.
YES you are expected to answer me :)