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The problem is asking to determine the sum of the possible values that \((a! + n)!\) can take, given the following conditions:

- \(a\) and \(n\) are natural numbers.
- \((n!)! = n\)
- \((a!)! = a + 1\)

Let’s first analyze each equation step by step:

### Step 1: Analyze \(n\)
From the condition \((n!)! = n\), we need to find \(n\) such that:
\[
(n!)! = n
\]
Trying small values of \(n\):
- For \(n = 1\), \((1!)! = 1! = 1\), which satisfies the equation.
- For \(n = 2\), \((2!)! = 2! = 2\), which also satisfies the equation.
- For \(n = 3\), \((3!)! = 6!\), but \(6! = 720\), and \((720)! \) is far too large to be equal to 3.

Thus, the only possible values of \(n\) are \(n = 1\) and \(n = 2\).

### Step 2: Analyze \(a\)
From the second condition \((a!)! = a + 1\), we need to find values of \(a\) such that:
\[
(a!)! = a + 1
\]
Trying small values of \(a\):
- For \(a = 1\), \((1!)! = 1! = 1\), and \(a + 1 = 2\). This does not satisfy the equation.
- For \(a = 2\), \((2!)! = 2! = 2\), and \(a + 1 = 3\). This does not satisfy the equation.
- For \(a = 3\), \((3!)! = 6!\), but \(6! = 720\), and clearly, \((720)! \) is far larger than \(a + 1 = 4\).

Thus, no natural number \(a\) satisfies \((a!)! = a + 1\) based on small trials.

Therefore, it seems there's a specific issue with the interpretation, and I'd need to reconsider the full analysis to find the correct logic behind \(a!\). Would you like further clarification on how we can approach this?

a ve n birer doğal sayıdır. (n!) = n ve (a!) = a + 1 olduğuna göre, (a! + n)! ifadesinin alabileceği değerler toplamı kaçtır?

Explore a factorial problem where you calculate the sum of possible values of (a! + n)! given the conditions (n!) = n and (a!) = a + 1. Learn how to work with factorials and natural numbers, suitable for Grades 9-10.

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