Starting with a positive integer $$M$$ written on the board, Alice plays the following game: in each move, if $$x$$ is the number on the board, she replaces it with $$3x+2$$. Similarly, starting with a positive integer $$N$$ written on the board, Bob plays the following game: in each move, if $$x$$ is the number on the board, he replaces it with $$2x+27$$. Given that Alice and Bob reach the same number after playing $$4$$ moves each, find the smallest value of $$M+N$$.

After each of Alice’s moves the number on the board is transformed by $$x \mapsto 3x+2$$.
Let $$A_n$$ be Alice’s number after $$n$$ moves, starting with $$A_0=M$$.

Repeated use of the transformation gives a geometric series:
$$A_1 = 3M+2$$
$$A_2 = 3(3M+2)+2 = 3^2M + 2(1+3)$$
Continuing, after $$n$$ moves,

$$A_n = 3^nM + 2\bigl(1+3+3^2+\dots+3^{\,n-1}\bigr)$$
The sum inside the brackets is a geometric sum:
$$1+3+\dots+3^{\,n-1}= \frac{3^{\,n}-1}{3-1}=\frac{3^{\,n}-1}{2}.$$ Therefore,

$$A_n = 3^nM + 2\cdot\frac{3^{\,n}-1}{2}=3^nM + (3^{\,n}-1).$$

For $$n=4$$, Alice reaches
$$A = 3^4M + (3^4-1)=81M+80.$$

Bob’s move is $$x \mapsto 2x+27$$. Let $$B_n$$ be his number after $$n$$ moves, starting with $$B_0=N$$.

Similar steps give
$$B_n = 2^nN + 27\bigl(1+2+\dots+2^{\,n-1}\bigr)$$
and since $$1+2+\dots+2^{\,n-1}=2^{\,n}-1,$$ $$B_n = 2^nN + 27(2^{\,n}-1).$$

For $$n=4$$, Bob reaches
$$B = 2^4N + 27(2^4-1)=16N+27\cdot15=16N+405.$$

Given that both reach the same number after four moves,

$$81M+80 = 16N+405.$$ Re-arranging,

$$81M - 16N = 325.$$

This is a linear Diophantine equation. Because $$\gcd(81,16)=1$$, solutions exist and can be parametrised.

Reduce the equation modulo $$16$$:
$$81M \equiv 325 \pmod{16}.$$
Since $$81\equiv1 \pmod{16}$$, we get
$$M \equiv 325 \equiv 5 \pmod{16}.$$

Write $$M = 5 + 16k$$ for some integer $$k$$. Substitute back:

$$81(5+16k)-16N = 325$$
$$405 + 1296k - 16N = 325$$
$$1296k + 80 = 16N$$
$$N = 5 + 81k.$$

Thus every positive solution is
$$M = 5 + 16k,\qquad N = 5 + 81k,\qquad k \in \mathbb{Z}.$$

To minimise $$M+N = 10 + 97k$$ we choose the smallest non-negative $$k$$, namely $$k=0$$.

Hence $$M=5,\;N=5$$ and the smallest value of $$M+N$$ is

10.