This question asks you to compute the sum of the digits of a factorial of number. Note that, factorial of a is denoted by n ! and compates to n(n1)321 n!=n(n1)(n2)+(3)(2)(1) For example, 10!=109321=3628800, and the sum of the digits in the number 10 t is 3+6+2+8+8+0+0=27 You are asked to write a function that takes an input number n and calculates the sam of all factorial digits. The first line of the function should read: function output = sum_factorial digits (n) The MATLAB code can build on following logic hint: - For a number n. the last digit is obtained as the remainder of dividing the number by 10 using command rem(n,10). Example: for n=7265, rem (n,10) will give us 5 . the second last digit is obtained by first dividing the number by 10 , taking the integer part of the quotient, and then finding its last digit as in the previots point. The MATL.AB command to get the second last digit is rem (f1 oor (n/10),10). Example: for n=7265, rem (f100r(n/10),10) will give us 6 . Likewise, the k-th last digit is given by dividing the number by 10k1, taking the integer part of the quotient, and finding the last digit in that. The MATLAB command to get the k-th last digit is rem (f100r(n/10(k1)),10). Examples for n=7265, to get the 42 last digit rem (fl00r(n/103),10) will give us 7 . - Hint \#2: Use a loop structure to loop through all digits that stops the process when the value of floor (n/10(k1) is 0 which means there are no more digits. - Hint \#3: Use factorial (n) function to calculate the value of factorial of n ( n !) As your submission include the following: - Complete M-file script. - Evaluated output of the function for following inputs: sum_factorial_digits (11) sum_factorial_digits (21)