Statistics of n dice adding up to 117 1. The problem statement, all variables and given/known data
(I translated this from french)
Some n dice are thrown and the sum of all their face values is 117.
Estimate n through a confidence interval of 90%. 
In other words, of all the possible rolls of n dice that sum up to 117, find a range in which n should be located 90% of the time.
For example, 20≤n≤117, 100% of the time because a 117-dice roll can sum up to 117, but not a 118-dice roll, and 19 dice can give a maximum sum of 114.
2. Relevant equations
To estimate the population average (here, the population is every possible roll of n dice that sums to 117) while the variance is unknown, and X is normally distributed (which it probably is), we use the following formula
$\frac{X^―−μ}{\sqrt{S^{2} _{n-1}/n}} : T_{n−1}$
Where :
X― is the sample average if that has anything to do with anything.
$S^{2} _{n-1}$ would be equal to $(n/n−1) S_2$where $S_2$ would be the sample's variance and n the number of items in the sample. 
Tn−1 is the symbol for Student's t-distribution.
There are five other, closely related, formulae, but this is the one that seems the most reasonable to use.
But I'm not even sure how that can be used to estimate n.
3. The attempt at a solution
What has been shown above probably shows at which point I am lost in this affair. We have to use some distribution, but I'm not even sure my choice is right.
 A: As I stated in a comment, the question makes no sense as posed since a confidence interval for just one value is meaningless; what is required to speak of a confidence interval is an entire prescription for determining confindence intervals from arbitrary observations. The formulation "find a range in which $n$ should be located $90\%$ of the time" seems to suggest that the person formulating the problem may misunderstand the concept of a confidence interval, as it seems to suggest a fixed interval which may or may not contain a varying $n$ $90\%$ of the time, rather than vice versa.
I doubt that you literally want an exact $90\%$ confidence interval; that would be a rather tedious calculation. I suspect that the idea is to approximate the distribution by a normal distribution. (No, it isn't one already.) A suitable $90\%$ confidence interval would then be one that includes all values of $n$ for which the observed sum $s$ is within approximately $1.645$ standard deviations of the mean. The mean of a single roll is $7/2$, and the variance of a single roll is $35/12$, so the mean is $\frac72n$ and the variance is $\frac{35}{12}n$. Thus the condition for $s$ to be within $1.645$ standard deviations of the mean is
$$
\left\vert s-\frac72n\right\vert\le1.645\sqrt{\frac{35}{12}n}\approx2.809\sqrt n\;.
$$
Squaring yields a quadratic equation for the interval boundaries:
$$
\left(s-\frac72n\right)^2\approx7.893n\;,\\
n^2-\frac47sn+\frac4{49}s^2-0.644n\approx0\;,\\
n\approx\frac27s+0.322\pm\sqrt{0.184s+0.104}\;.
$$
For $s=117$, this yields $n\approx33.751\pm4.651$, corresponding to a confidence interval of approximately $[29.1,38.4]$. Thus, with $90\%$ confidence $n$ lies betwen $30$ and $38$.
