Add subsequent natural numbers with a max difference of a prime to get to 119 I found a really interesting math challenge that was given to 7 graders (around 13-14 years old). It goes like this:
Problem
Anne added subsequent natural numbers to get to $119$. The difference between the highest and lowest number is prime. 
Find the subsequent numbers that fulfill this condition and show that this is the only solution.
Attempt at the solution
By factoring into primes, $119$ becomes $17\cdot 7$. Utilizing an adapted Gauss formula of sums
$$ \sum_{i=n_l}^{n_u} i = \frac{n_u-n_l + 1}{2}\cdot (n_u+n_l),$$
we can look for $7$ numbers with a sum of the lowest and highest number of $17$ to find $n_u=n_l+6$ and thus
$$ 119 = \sum_{i=n_l}^{n_l + 6} = \frac{7}{2}\cdot (2n_l+6)=7n_l+21 $$
$$\Rightarrow n_l=14$$
This way, one possible solution is $14+15+16+17+18+19+20$, but the difference is not prime. 
After a lot of rewriting the formula I found the solution
$$2+3+4+5+6+7+8+9+10+11+12+13+14+15$$
with 13 as a prime difference, but in no way I can show that this is the only solution. Also, how is a 7 grader supposed to do this without knowing what a variable is? 
Question
Are there any easy solutions I missed? How can you utilize the prime condition?
 A: First off, you know that the number of naturals summed must be even, since the difference must be odd. Moreover, the even must be $p+1$ for some prime $p$. For a set of $2n$ consecutive naturals to sum up to $119$, $\frac{119}{2n}+\frac12$ must be an integer (since $(x-n)+(x-n+1)+...+(x+n+1)=2nx+n$). Hence, $n$ must divide $119$. So, we want to find a number $2n-1$ such that $2n-1$ is prime and $n\mid119$.
Since $119=7\cdot17$, the only choices for $n$ that we have are $1,7,17,119$. Clearly, $119$ doesn't work, and neither does $1$. 
$2\cdot7-1=13$ works, but $2\cdot17-1=33$ doesn't, since it is not prime. So, $13$ is the only possibility.
I don't know how to do this without variables. Maybe use intuition with some of the formulas here?
A: This is a seventh-grader solution.
Let each number be represented by a row of counters. Now arrange the counters into a trapezium,  Let $h$ be the height of the trapezium (how many numbers there are), $a$ be the top base (the number of counters at the top), and $b$ be the bottom base (the number of counters on the bottom).

Then we have that:
$$\frac{1}{2}(a+b)h = 7\cdot17$$
$$(a+b)h = 2 \cdot 7 \cdot 17$$
Then they can test values of different values of $h$. To start off, in order for $a+b$ to be an integer, the only possible values of $h$ are the factors of this number: $1,2,7,14,17,34,119$ and $238$. In addition, the prime difference is one less than how many numbers there are. Students can reason this by writing down the list of numbers in one row, and counting the number of numbers in another row.
Subtracting $1$ from each value leaves the only candidate as $14$, as the rest are either divisible by $2$ or by $3$. Since there are $14$ numbers in total, the difference between the largest and smallest numbers is $14-1=13$.
