3 in 1

For the first problem, we use the property $a.b=gcd(a,b).lcm(a,b)$ .

The above property tells us that at any stage, the blackboard contains at least one even number, because the parity of the product of all the numbers on the board is invariant, namely it is even.

To show the invariance, we check that the product of all the numbers is even at first because of the presence of $2$ in the set initially.

Now if in all the steps we do our operations with the odd numbers only then the $2$ will always remain, making the product even, and if the $2$ is taken anytime for the operation, the above property preserves the parity of the product of the two numbers which is even on L.H.S. due to presence of $2$ so R.H.S. must also be even, so at least one of the factors on R.H.S. is even.

Hence the final product is also even. So there exists at least one even number in the blackboard at all times. Hence $Y=10$ .

For the second problem,

we make the following observations:

Each time the man makes a step of first kind (as in question) we write a $1$ , each time he makes a step of the second kind, we write $\omega$ where $\omega$ is the third root of unity, and each time he makes a step of the third kind, we write ${\omega}^2$ which is the other third root of unity, so that the net displacement of the man in terms of complex numbers is $a+b\omega+c{\omega}^2=1$ since the net displacement is 1. Also $a+b+c=10$ . Also note that the net displacement is only in the horizontal direction (that is along x-axis) so for each $\omega$ step we have to take an ${\omega}^2$ step. So we get $b=c$ .

Thus the two equations give $a+2b=10$ and $a-b=1$ . Solving the two equations we get $S=$ { $4,3$ }, so that $X=7$ .

For the third problem, the second condition gives us the hint that $f(m,n)$ is actually the G.C.D of $m$ and $n$ . We take the help of Euclid's algorithm for finding out the G.C.D. of the two integers $m,n$ . We note that the second condition, can also be viewed as $f(p,q)=f(p,p-q)$ (if $p>q$ ) (this can be assumed without loss of generality due to the first condition that $f(m,n)=f(n,m)$ ) by simply putting $p=m+n,q=n$ . So $f(p,q)=f(p,p-q)=f(p,p-2q)=....=f(p,p-aq)$ where a is the largest integer such that $p-aq>0$ . So $a$ must be the quotient on dividing $p$ by $q$ so that $p-aq$ is actually the remainder of the said division. So we have actually found that

$f(p,q)=f(p,p(mod(q)))$ and by the Euclidean algorithm, we know that this process terminates of taking the dividend and the remainder terminates when the remainder is zero. In that step the divisor is actually the G.C.D of the two numbers $p,q$ we started with.

$f(p,q)=f(gcd(p,q),0)=gcd(p,q)$ where the last equality follows from the third condition.

Hence $f(m,n)=gcd(m,n)$ . So $Z=29$ .

Which gives Ramanujan's Number $1729$ as the answer.