Deceptively Complex

The idea is to break down this process into several subprocesses.

Take a look at the bottom right corner. It keeps subtracting 17 from B when B is larger than 17, this will in the end return $B\mod 17$ , except when the result is 0, it will return 17.

Now, let's look at the process that starts from $\text{Set Counter to 0}$ . The process after this will be repeated $A$ times. More precisely, we perform $\text{Add B to B}$ for $A$ times. $\text{Add B to B}$ is another way of saying $\text{Multiply B by 2}$ . So the entire process will multiply $B$ by $2^A$ .

Again, the whole process keeps repeating for $A=6,5,4,3,2,1$ . Hence, B is multiplied by 2 for $5+4+3+2+1+0$ times. B becomes

$B \times 2^5 \times 2^4\times 2^3 \times 2^2 \times 2^1 = B \times 2^{15} = 2^2 \times 2^{15} = 2^{17}$

Recall that we keep subtracting $17$ from B, and we know that $a^n\bmod b = (a\bmod b)^n \bmod b$ , so it doesn't matter if we do the modulo first or multiplication first. Finally, by Fermat's Little Theorem ,

$2^{17}\bmod {17} = \boxed2.$

Here is my Python solution, since Christopher stole my solution :P.

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def F(a,b):
    while a>0:
        a=a-1
        counter=0
        while a!=counter:
            counter+=1:
            b=(2*b)%17
    if a=0:
        return b

print (F(6,4))

This outputs $2$ as the answer. Note: If the output is $0$ , it should be interpreted as $17$ .