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I am not sure how to find an answer to this question. Is there a way to solve it without simply trial and error?
Do there exist ten distinct positive integers such that their arithmetic mean is
(a) six times as great as their greatest common divisor?
(b) five times as great as their greatest common divisor?
Let us call the ten positive integers $a_1 , a_2 , \dots , a_{10}$ and let us call their GCD $d$. Note that $d \mid a_i$ for each $i$. We can write $a_i = b_i d $ with $b_i$ a positive integer.
The condition for a) is: $$\frac{a_1 + a_2 + \dots + a_{10}}{10} = 6 d$$
Let us divide by $d$ to get:
$$\frac{b_1 + b_2 + \dots + b_{10}}{10} = 6 $$
So the question turns into are their $10$ distinct positive integers, $b_i$ such that their arithmetic mean is $6$, equivalently $$b_1 + b_2 + \dots + b_{10} = 60 $$ are their $10$ distinct positive integers $b_i$ such that their sum is $60$. (To be precise we should also say whose GCD is $1$ but this is a detail).
Recall that the sum of the first $10$ positive integers is already $1+ 2 + \dots + 10 = 55$. So let us start $b_1 = 1, b_2 = 2, \dots, b_9 = 9$ and then $b_10= 15$.
You can now choose $d$ to be whatever you like and multiply $b_i$ by it to get the $a_i$.
If you want to make it simple just take $d=1$ and you are done.
For b) it would be that the sum must be $50$. This is not possible.
