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Difference between revisions of "2023 AMC 12B Problems/Problem 23"

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==Solution==
  
 +
The product can be written as
 +
<cmath>
 +
\begin{align*}
 +
2^a 3^b 4^c 5^d 6^e
 +
& = 2^{a + 2c + e} 3^{b + e} 5^d .
 +
\end{align*}
 +
</cmath>
 +
 +
Therefore, we need to find the number of ordered tuples <math>\left( a + 2c + e, b+e, d \right)</math> where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math> are non-negative integers satisfying <math>a+b+c+d+e \leq n</math>.
 +
We denote this number as <math>f(n)</math>.
 +
 +
Denote by <math>g \left( k \right)</math> the number of ordered tuples <math>\left( a + 2c + e, b+e \right)</math> where <math>\left( a, b, c, e \right) \in \Delta_k</math> with <math>\Delta_k \triangleq \left\{ (a,b,c,e) \in \Bbb Z_+^4: a+b+c+e \leq k \right\}</math>.
 +
 +
Thus,
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
Next, we compute <math>g \left( k \right)</math>.
 +
 +
Denote <math>i = b + e</math>. Thus, for each given <math>i</math>, the range of <math>a + 2c + e</math> is from 0 to <math>2 k - i</math>.
 +
Thus, the number of <math>\left( a + 2c + e, b + e \right)</math> is
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
Therefore,
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
By solving <math>f \left( n \right) = 936</math>, we get<math>n = \boxed{\textbf{(A) 11}}</math>.
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 17:41, 15 November 2023

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)

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