\[ 3 \times 4 = 3+3+3+3=4+4+4 \] \[ n!=n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1 \]

Finding the factorial of any natural number is just finding the product of first \(n\) natural numbers, moreover product of two positive numbers \(m\) and \(n\) is just the summation of \(m\) \(n's\) or \(n\) \(m's\). Therefore the factorial of a number can also be calculated using only the addition operator.

**For example:**

\( 0!=1 \) (No addition)

\( 1!=1 \) (No addition)

\( 2!=2 \) (No addition)

\(3!=3 \times 2! \) OR \(3!=2!+2!+2!\) (2 additions)

\(4!=4 \times 3! \) OR \(4!=3!+3!+3!+3!\) (How many addition here? Do you see a pattern?)

What is the minimum number of additions that you need to perform to find the factorial of \(10\) when you are allowed to use only (+) operation ? (**Hint:** Generalize it for the factorial of any non-negative integer \(N\))

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