Hypothetical Factorial Question Solution

Hi, I'm new to MATLAB and I'm teaching myself. I have made this code to answer what is the factorial of 'n'. My question is - is it possible to write code that will solve this backwards? i.e. if I'm given 120 can it solve for the answer of 5? Thanks in advance!

n = 5;
nfact = 1;
for i = 1:n
    nfact = nfact*i;
end
disp(nfact)

[David Goodmanson]

HI Deborah,

One straightforward way would be: divide by 2. If that result is an integer, divide it by 3. If that result is an integer, divide it by 4 ... keep going until the division either returns 1 somewhere along the line, in which case the starting point was the factorial of the last-used integer, or does not return 1 and returns values less than 1, in which case the starting point was not the factorial of an integer. Programming that up should give you more good practice with Matlab.

If you go this direction you will of course have to check if a number is an integer. Matworks has an 'isinteger' function, but somewhat perversely it does not work for ordinary Matlab double precision numbers. But you can use

floor(n) == n

and if the result is true, n is an integer.

Answer 1

[Chunru]

x = 120;
nfact = 1;
for i = 1:x
    nfact = nfact*i;
    if nfact==x
        fprintf("%d = %d!\n", x, i)
        break
    elseif nfact>x
        fprintf('x is not a factorial number. \n')
        break;
    end
end

120 = 5!

[the cyclist]

You might want to be careful about floating-point error in the equality check, and instead use something like

tol = 1.e-6;
if abs(nfact-x) < tol

[Chunru]

This is supposed to be integer operation of factorial.

Answer 2

[the cyclist]

I don't really intend this to be a serious answer to your question, but according to the internet, this formula will work up to N==170.

% Original number
N = 170;

% Factorial of the number
nfact = factorial(N);

% Recover original number
n = round((log(nfact))^((303*nfact^0.000013)/(144*exp(1))) + 203/111 - 1/nthroot(nfact,exp(1)))

n = 170

Looks like this is based on Stirling's approximation.

Also, note that 171! is too large to be represented as double precision in MATLAB:

factorial(170)

ans = 7.2574e+306

factorial(171)

ans = Inf

[Stephen23]

Even 23! is too large to be accurately represented using the DOUBLE class:

factorial(sym(23))

ans = 

25852016738884976640000

sym(factorial(23))

ans = 

25852016738884978212864