smc

We start by trying to prove a function of $n$, and then we can apply the function and equate it to $936$ to find the value of $n$. It is helpful to think of this problem in the format $(1+2+3+4+5+6)\cdot (1+2+3+4+5+6)\dots$. Note that if we represent the scenario in this manner, we can think of picking a $1$ for one factor and then a $5$ for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for $n=2$, $4$ can be reached by picking $1$ and $4$ or $2$ and $2$. However, this form gives us insights that will be useful later in the problem. Note that there are only $3$ primes in the set $\{1,2,3,4,5,6\}$: $2,3,$ and $5$. Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form $2^h\cdot 3^i\cdot 5^j$ (the choice of variables will become clear later), for integer nonnegative values $h,i,j$. So now the problem boils down to how many distinct triplets $(h,i,j)$ can be formed by taking the product of $n$ dice values. We start our work on representing $j$: the powers of $5$, because it is the simplest in this scenario because there is only one factor of $5$ in the set. Because of this, having $j$ fives in our prime factorization of the product is equivalent to picking $j$ factors from the polynomial $(1+\cdots +6)\cdots$ and choosing each factor to be a $5$. Now that we've selected $j$ factors, there are $n-j$ factors remaining to choose our powers of $3$ and $2$. Suppose our prime factorization of this product contains $i$ powers of $3$. These powers of $3$ can either come from a $3$ factor or a $6$ factor, but since both $3$ and $6$ contain only one power of $3$, this means that a product with $i$ powers of $3$ corresponds directly to picking $i$ factors from the polynomial, each of which is either $3$ or $6$ (but this distinction doesn't matter when we consider only the powers of $3$. Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair $(i,j)$ that match the requirements, corresponding to the number of $3$'s and the number of $5$'s our product will have. Then how many different $h$ values for the powers of $2$ are possible? In the $i+j$ factors we have already chosen, we obviously can't have any factors of $2$ in the $j$ factors with $5$. However, we can have a factor of $2$ pairing with factors of $3$, if we choose a $6$. The maximal possible power of $2$ in these $i$ factors is thus $2^i$, which occurs when we pick every factor to be $6$. We now have $n-i-j$ factors remaining, and we want to allocate these to solely powers of $2$. For each of these factors, we can choose either a $1,2,$ or $4$. Therefore the maximal power of $2$ achieved in these factors is when we pick $4$ for all of them, which is equivalent to $2^{2\cdot (n-i-j)}$. Now if we multiply this across the total $n$ factors (or $n$ dice) we have a total of $2^{2n-2i-2j}\cdot 2^i=2^{2n-i-2j}$, which is the maximal power of $2$ attainable in the product for a pair $(i,j)$. Now note that every power of $2$ below this power is attainable: we can simply just take away a power of $2$ from an existing factor by dividing by $2$. Therefore the powers of $2$, and thus the $h$ value ranges from $h=0$ to $h=2n-i-2j$, so there are a total of $2n+1-i-2j$ distinct values for $h$ for a given pair $(i,j)$. Now to find the total number of distinct triplets, we must sum this across all possible $i$s and $j$s. Lets take note of our restrictions on $i,j$: the only restriction is that $i+j\le n$, since we're picking factors from $n$ dice. $$\sum _{i+j\le n}2n+1-i-2j=\sum _{i+j\le n}2n+1-\sum _{i+j\le n}i+2j$$ We start by calculating the first term. $2n+1$ is constant, so we just need to find out how many pairs there are such that $i+j\le n$. Set $i$ to $0$: $j$ can range from $0$ to $n$, then set $i$ to $1$: $j$ can range from $0$ to $n-1$, etc. The total number of pairs is thus $n+1+n+n-1+\cdots +1=\frac{(n+1)(n+2)}{2}$. Therefore the left summation evaluates to $$\frac{(2n+1)(n+1)(n+2)}{2}$$ Now we calculate ${\sum }_{i+j\le n}^{}i+2j$. This simplifies to ${\sum }_{i+j\le n}^{}i+2\cdot {\sum }_{i+j\le n}^{}j$. Note that because $i+j=n$ is symmetric with respect to $i,j$, the sum of $i$ in all of the pairs will be equal to the sum of $j$ in all of the pairs. Thus this is equal to calculating $3\cdot {\sum }_{i+j\le n}^{}i$. In the pairs, $i=1$ appears for $j$ ranging between $0$ and $n-1$ so the sum here is $1\cdot (n)$. Similarly $i=2$ appears for $j$ ranging from $0$ to $n-2$, so the sum is $2\cdot (n-1)$. If we continue the pattern, the sum overall is $(n)+2\cdot (n-1)+3\cdot (n-2)+\cdots +(n)\cdot 1$. We can rearrange this as $((n)+(n-1)+\cdots +1)+((n-1)+(n-2)+\cdots +1)+((n-2)+(n-3)+\cdots +1)+\cdots +1)$ $$=\frac{(n)(n+1)}{2}+\frac{(n-1)(n)}{2}+\cdots +1$$ We can write this in easier terms as ${\sum }_{k=0}^n\frac{(k)(k+1)}{2}=\frac{1}{2}\cdot {\sum }_{k=0}^n{k}^2+k$ $$=\frac{1}{2}\cdot (\sum _{k=0}^n{k}^2+\sum _{k=0}^{n}k)$$ $$=\frac{1}{2}\cdot (\frac{(n)(n+1)(2n+1)}{6}+\frac{(n)(n+1)}{2})$$ $$=\frac{1}{2}\cdot (\frac{(n)(n+1)(2n+1)}{6}+\frac{3n(n+1)}{6})=\frac{1}{2}\cdot \frac{n(n+1)(2n+4)}{6}$$ $$=\frac{n(n+1)(n+2)}{6}$$ We multiply this by $3$ to obtain that $$\sum _{i+j\le n}i+2j=\frac{n(n+1)(n+2)}{2}$$ Thus our final answer for the number of distinct triplets $(h,i,j)$ is: $$\sum _{i+j\le n}2n+1-i-2j=\frac{(2n+1)(n+1)(n+2)}{2}-\frac{n(n+1)(n+2)}{2}$$ $$=\frac{(n+1)(n+2)}{2}\cdot (2n+1-n)=\frac{(n+1)(n+2)}{2}\cdot (n+1)$$ $$=\frac{(n+1{)}^2(n+2)}{2}$$ Now most of the work is done. We set this equal to $936$ and prime factorize. $936=12\cdot 78=2^3\cdot 3^2\cdot 13$, so $(n+1{)}^2(n+2)=936\cdot 2=2^4\cdot 3^2\cdot 13$. Clearly $13$ cannot be anything squared and $2^4\cdot 3^2$ is a perfect square, so $n+2=13$ and $n=11=\boxed{11}$