# Use only '+' operator

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

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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