2023 AMC 12B Problems/Problem 23

Solution

The product can be written as $\begin{align*} 2^a 3^b 4^c 5^d 6^e & = 2^{a + 2c + e} 3^{b + e} 5^d . \end{align*}$

Therefore, we need to find the number of ordered tuples $\left( a + 2c + e, b+e, d \right)$ where $a$ , $b$ , $c$ , $d$ , $e$ are non-negative integers satisfying $a+b+c+d+e \leq n$ . We denote this number as $f(n)$ .

Denote by $g \left( k \right)$ the number of ordered tuples $\left( a + 2c + e, b+e \right)$ where $\left( a, b, c, e \right) \in \Delta_k$ with $\Delta_k \triangleq \left\{ (a,b,c,e) \in \Bbb Z_+^4: a+b+c+e \leq k \right\}$ .

Thus, $\begin{align*} f \left( n \right) & = \sum_{d = 0}^n g \left( n - d \right) \\ & = \sum_{k = 0}^n g \left( k \right) . \end{align*}$

Next, we compute $g \left( k \right)$ .

Denote $i = b + e$ . Thus, for each given $i$ , the range of $a + 2c + e$ is from 0 to $2 k - i$ . Thus, the number of $\left( a + 2c + e, b + e \right)$ is $\begin{align*} g \left( k \right) & = \sum_{i=0}^k \left( 2 k - i + 1 \right) \\ & = \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) . \end{align*}$

Therefore, $\begin{align*} f \left( n \right) & = \sum_{k = 0}^n g \left( k \right) \\ & = \sum_{k=0}^n \frac{1}{2} \left( k + 1 \right) \left( 3 k + 2 \right) \\ & = \frac{3}{2} \sum_{k=0}^n \left( k + 1 \right)^2 - \frac{1}{2} \sum_{k=0}^n \left( k + 1 \right) \\ & = \frac{3}{2} \cdot \frac{1}{6} \left( n+1 \right) \left( n+2 \right) \left( 2n + 3 \right) - \frac{1}{2} \cdot \frac{1}{2} \left( n + 1 \right) \left( n + 2 \right) \\ & = \frac{1}{2} \left( n + 1 \right)^2 \left( n + 2 \right) . \end{align*}$

By solving $f \left( n \right) = 936$ , we get $n = \boxed{\textbf{(A) 11}}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Page

Toolbox

Search

2023 AMC 12B Problems/Problem 23

Solution