2023 AMC 12B — Problem 23
Problem Statement
When a standard \(6\)-sided die is rolled \(n\) times, the product of the \(n\) numbers rolled can be any of \(936\) possible values. What is \(n\)?
Complete Analytical Solution
Step 1: Mapping the Product to Prime Exponents
A standard \(6\)-sided die has the face values \(\{1, 2, 3, 4, 5, 6\}\). Let \(x_1, x_2, x_3, x_4, x_5, x_6\) denote the number of times each face is rolled across \(n\) total rolls. These counts must satisfy the non-negative integer constraint:
\(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=n\)
The total product of the \(n\) outcomes is given by:
\(\text{Product}=1^{x_{1}}\cdot 2^{x_{2}}\cdot 3^{x_{3}}\cdot 4^{x_{4}}\cdot 5^{x_{5}}\cdot 6^{x_{6}}\)
By breaking the composite faces down into their prime factors (\(4 = 2^2\) and \(6 = 2 \times 3\)), we can regroup the product strictly in terms of the prime bases \(2, 3,\) and \(5\):
\(\text{Product}=2^{(x_{2}+2x_{4}+x_{6})}\cdot 3^{(x_{3}+x_{6})}\cdot 5^{x_{5}}\)
Let us define the total prime exponents as \(A, B,\) and \(C\):
- \(A = x_2 + 2x_4 + x_6\) (The total exponent of \(2\))
- \(B = x_3 + x_6\) (The total exponent of \(3\))
- \(C = x_5\) (The total exponent of \(5\))
Every unique product value corresponds directly and uniquely to a valid triplet of integer exponents \((A, B, C)\). Therefore, the problem reduces to finding the number of distinct valid triplets \((A, B, C)\) under the constraint of \(n\) rolls.
Step 2: Fixing Independent Variables (Dimensional Reduction)
Because face \(5\) is the only face that generates the prime factor \(5\), the variable \(C = x_5\) is completely independent of \(A\) and \(B\). Let us fix \(C\).
The remaining faces \(\{1, 2, 3, 4, 6\}\) must account for the remaining rolls. Let \(m = n – C\) represent this remaining pool of rolls, yielding:
\(x_{1}+x_{2}+x_{3}+x_{4}+x_{6}=m\)
Next, we fix \(B = x_3 + x_6\), which represents the total exponent of \(3\). Since these rolls must come from faces \(3\) and \(6\), the value of \(B\) can range anywhere from \(0\) to \(m\).
Once \(B\) is fixed, we can choose the number of \(6\)s rolled, \(x_{6}\), to be any integer such that \(0 \le x_6 \le B\). Choosing \(x_{6}\) automatically locks in the number of \(3\)s rolled via \(x_3 = B – x_6\). This leaves the remaining rolls to be distributed among faces \(1, 2,\) and \(4\):
\(x_{1}+x_{2}+x_{4}=m-B\)
Step 3: Determining the Boundary Ranges of Exponent \(A\)
To isolate the behavior of exponent \(A\), we eliminate \(x_{2}\) by substituting \(x_2 = (m – B) – x_1 – x_4\) into its formula:
\(A=x_{2}+2x_{4}+x_{6}=\left[(m-B)-x_{1}-x_{4}\right]+2x_{4}+x_{6}\)
\(A=(m-B)-x_{1}+x_{4}+x_{6}\)
For fixed values of \(m, B,\) and \(x_{6}\), the value of \(A\) varies based entirely on how we allocate rolls between \(x_{1}\) and \(x_{4}\):
- Minimum value of \(A\): Achieved when we maximize \(x_{1}\) (letting \(x_1 = m – B, x_4 = 0\)):
\(A_{\min }=x_{6}\) - Maximum value of \(A\): Achieved when we maximize \(x_{4}\) (letting \(x_1 = 0, x_4 = m – B\)):
\(A_{\max }=2(m-B)+x_{6}\)
Because we can systematically swap out a \(1\) for a \(2\), or a \(2\) for a \(4\), the value of \(A\) can increase incrementally by exactly \(1\) at each step. Thus, \(A\) can take any integer value in the continuous interval \([x_6, 2m – 2B + x_6]\).
Taking the union of these intervals across all valid choices of \(x_{6}\) (from \(0\) to \(B\)) yields the absolute baseline range for \(A\) given a fixed \(B\):
\(0\le A\le 2m-B\)
Thus, for any chosen value of \(B\), the number of unique values that \(A\) can take is exactly:
\(\text{Count}(A)=(2m-B)-0+1=2m-B+1\)
Step 4: Summation Over All Cases
To find the total number of distinct products for a fixed \(m\), we sum over all possible values of \(B\) from \(0\) to \(m\):
\(\sum _{B=0}^{m}(2m-B+1)\)
This forms a simple arithmetic progression: \((2m+1) + 2m + (2m-1) + \dots + (m+1)\). Using the arithmetic series sum formula \(\frac{(\text{First}+\text{Last})\times \text{Terms}}{2}\):
\(\text{Combinations\ per\ }m=\frac{(2m+1)+(m+1)}{2}\times (m+1)=\frac{(3m+2)(m+1)}{2}=\frac{3m^{2}+5m+2}{2}\)
Finally, we sum this quadratic expression across all valid values of \(m\) from \(0\) to \(n\) to account for all possible values of \(C\):
\(\text{Total\ Products}=\sum _{m=0}^{n}\frac{3m^{2}+5m+2}{2}\)
Using the standard power summation formulas (\(\sum 1 = n+1\), \(\sum m = \frac{n(n+1)}{2}\), and \(\sum m^2 = \frac{n(n+1)(2n+1)}{6}\)), this factors out cleanly into the closed-form Factor Exponent Sum Formula:
\(\text{Total\ Products}=\frac{(n+1)^{2}(n+2)}{2}\)
Step 5: Solving for \(n\) via Structural Factorization
We are given that the total number of unique products is \(936\). Setting our closed-form general formula equal to this target:
\(\frac{(n+1)^{2}(n+2)}{2}=936\)
Multiplying both sides by 2 to clear the fraction yields:
\((n+1)^{2}(n+2)=1872\)
To solve this equation rigorously without expanding it into a cubic polynomial, we perform a prime factorization of the number \(936\) to break down its internal arithmetic structure:
\(936=8\times 117=2^{3}\times 3^{2}\times 13\)
Now, multiplying it by 2 to match the value of our target integer \(1872\):
\(1872=2\times (2^{3}\times 3^{2}\times 13)=2^{4}\times 3^{2}\times 13\)
We strategically regroup these prime factors into a product consisting of a perfect square and a baseline integer, mimicking the algebraic format of \((n+1)^2(n+2)\):
\(1872=(2^{4}\times 3^{2})\times 13=(16\times 9)\times 13=144\times 13\)
Since \(144 = 12^2\), we substitute it back to directly align our structural variables:
\((n+1)^{2}(n+2)=12^{2}\times 13\)
By direct matching of the algebraic components:
- \(n + 1 = 12 \implies n = 11\)
- \(n + 2 = 13 \implies n = 11\)
Final Answer
The number of die rolls is \(n = 11\).
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