The sum of $49$ natural numbers is $540$. Find the largest possible value of their greatest common divisor. 
The sum of $49$ natural numbers is $540$. Find the largest possible value of their greatest common divisor.

I don't really understand even how the proof should be structured here. It must be shown that the gcd of numbers does not exceed some natural number $d'$, right? Is it going to be enough? I'd be thankful if you could show me a full, formal proof.
 A: Denote the numbers by $x_1, x_2, \ldots, x_{49}$ and their greatest common divisor by $g$.  Then $g \le x_i$ for each $i$, and so $g \le \min(x_1, \ldots, x_{49})$.  But the minimum is less than or equal to the average of the numbers, so $g \le 540/49$.  Since $g$ is an integer we have $g \le 11$.
Next, $540 = x_1 + x_2 + \cdots + x_{49}$.  Let $x_i = g y_i$ for each $i$; the $y_i$ are positive integers because $g$ is a divisor of $x_i$.  So $540 = g(y_1 + \cdots + y_{49})$ and therefore $g$ is a factor of $540$.
So $g$ can't be $11$.  It can be $10$, and we can construct an explicit example, $x_1 = x_2 = \cdots = x_{48} = 10$ and $x_{49} = 60$.  Similarly $g$ can be any factor of $540$ smaller than 11.
A: We are trying to satisfy the equation
$$\sum_{i=1}^n a_ix_i=540$$
where $\sum_{i=1}^n a_i=49$ and $x_i\in\Bbb N$.
I propose that the answer is $10$, since we have $$540=48\times 10+1\times 60 \Rightarrow\gcd(10,60)=10$$
(as per my comment above).
Suppose the answer is greater than $10$.  We know that the maximum $\gcd$ of a set is at most equal to one of the elements in the set.  In order to produce a $\gcd$ greater than $10$ (say, $11$), then we must have another element in the set that is a multiple of that number (i.e. $11k, k\in\Bbb N, k>1$).  We would then have
$$a_1(11)+a_2(11k)=540.$$
Notice that this does not work as  $540\equiv 1\pmod{11}$, or more specifically, $540\not\equiv 0\pmod{11}$.
So, in addition to a potential $\gcd$ value greater than $10$, it must also divide $540$.  The next lowest factor of $540$ is $12$, so we can try as we had above:
\begin{align}
a_1(12)+a_2(12k)&=540\\
12(a_1+ka_2)&=540\\
a_1+ka_2&=45
\end{align}
But from our original equation, we need $a_1+a_2=49$.  So, the $\gcd$ you are looking for cannot be $12$, nor can it be greater than $12$.  Therefore, the answer must be $10$.
A: It cannot be $11$ (as mentioned above).  So... 
$$10$$
$$44 \cdot 10 + 5 \cdot 20 = 540$$
(or similar answers by trading $10$s for multiples of $10$)
