Arithmetic Puzzle The problem is as follows:
We start with the number $\mathbf 1$. The puzzle consists of $6$ steps. At each step, we can choose to do an operation on the resulting number we have at that step. The goal of the puzzle is to somehow reach the number $\mathbf{49}$. Each operation must be used exactly once. Here are the permitted operations we can do:
$\bullet \times 2$
$\bullet \times 3$
$\bullet \times 5$
$\bullet +2$
$\bullet +3$
$\bullet +5$
Now this might seem like an easy problem to you (maybe it is), but I haven’t been able to do any good. I tried wildly guessing ways to do this in vain. I certainly don’t intend to check all $6!$ ways (lesser than that, but still too many).
Here is a valid combination as an example:
$$1 \overset{\times 5}{\longrightarrow} 5 \overset{+2}{\longrightarrow} 7  \overset{\times 3}{\longrightarrow} 21  \overset{\times 2}{\longrightarrow} 42 \overset{+5}{\longrightarrow} 47  \overset{+3}{\longrightarrow} 50$$
Can anyone spot a solution to this puzzle? And more importantly, is there an analytical way to go about this?

The last step must be an addition because $49$ is not divisible by $2,3,5$.
If we perform several additions or several multiplications in sequence, we may agree to perform them as one (e.g., as $+5,+3$ as $+8$ or $\times 2,\times 3$ as $\times 6$)
So we end with one of $+2,+3,+5,+7,+8,+10$, i.e., we cam by multiplication to one of $47,46,44,42,41,39$. By looking at feasible factors (in particular, ruling out the primes), the only possible paths are
$$ 49\stackrel{+3}\leftarrow 46\stackrel{\times 2}\leftarrow 23\\
49\stackrel{+5}\leftarrow 44\stackrel{\times 2}\leftarrow 22\\
49\stackrel{+7}\leftarrow 42\stackrel{\times 6}\leftarrow 7\\
49\stackrel{+7}\leftarrow 42\stackrel{\times 3}\leftarrow 14\\
49\stackrel{+7}\leftarrow 42\stackrel{\times 2}\leftarrow 21\\
49\stackrel{+10}\leftarrow 39\stackrel{\times 3}\leftarrow 13
$$
In the last case, we used up all additions and have lost.
In rows three to five, the only addition left is $+3$, but $4, 11, 18$ are out of reach by multiplication alone.
So prepending the next (cluster of) additions, we only need to deal with (more  forced operations also prepended)
$$ 
49\stackrel{+3}\leftarrow 46\stackrel{\times 2}\leftarrow 23\stackrel{+2}\leftarrow21\stackrel{\times3}\leftarrow7\stackrel{+5}\leftarrow2\\
49\stackrel{+3}\leftarrow 46\stackrel{\times 2}\leftarrow 23\stackrel{+5}\leftarrow18\stackrel{\times3}\leftarrow6\stackrel{+2}\leftarrow4\\
49\stackrel{+3}\leftarrow 46\stackrel{\times 2}\leftarrow 23\stackrel{+7}\leftarrow16\\
49\stackrel{+5}\leftarrow 44\stackrel{\times 2}\leftarrow 22\stackrel{+2}\leftarrow 20\stackrel{\times 5}\leftarrow 4\stackrel{+3}\leftarrow1\\
49\stackrel{+5}\leftarrow 44\stackrel{\times 2}\leftarrow 22\stackrel{+3}\leftarrow 19\\
49\stackrel{+5}\leftarrow 44\stackrel{\times 2}\leftarrow 22\stackrel{+5}\leftarrow 17\\
$$
All dead ends ...
