Csdrhrt

I have a problem understand this math problem:

Write a number on the board. This number is either multiplied by $2$ or raised to a square. If the number $"1"$ is written at the beginning, at least how many steps should be taken to reach $2^{2018}?$

A) $15$

B) $16$

C) $17$

D) $18$

E) $12$

I can't solve this problem. Because I don't understand the question. Now, I need to understand the question. Then maybe I can.

Is there a problem in the question? The question unclear for me...Can you explain me, what is the meaning of the question?

Since you want to go to $2^{2018}$, it's easier to use exponents: the operations are doubling or adding $1$, start from $0$ and get to $2018$.

The best strategy is going backwards:
$$
2018 xrightarrow{/2} 1009 xrightarrow{-1} 1008
xrightarrow{/2} 504 xrightarrow{/2} 252 xrightarrow{/2} 126
xrightarrow{/2} 63 xrightarrow{-1} 62 xrightarrow{/2} 31
xrightarrow{-1} 30 xrightarrow{/2} 15 xrightarrow{-1} 14
xrightarrow{/2} 7 xrightarrow{-1} 6 xrightarrow{/2} 3
xrightarrow{-1} 2 xrightarrow{/2} 1 xrightarrow{-1} 0
$$
Is this the most efficient way?

At odd numbers you have no choice. Suppose the number $n$ is even; if it is a multiple of $4$, $n=4k$, you get at $k$ in two steps and this is obviously the best choice: otherwise you have to subtract $1$ twice, divide by $2$, subtract $1$ and divide again by $2$ just to arrive at $k-1$, five steps against two for a very small gain; in the average, this method is worse. If $n=4k+2$ you can choose between dividing by $2$ and subtracting $1$ or subtracting $1$ twice and dividing by $2$: two steps in the former case, three in the latter.

I think the problem means:

Given the two functions
$$ f(x) = 2x \g(x) = x^2 $$
consider sequences of $f$s and $g$s such that,
$$ g(f(g(g(cdots(g(f(1)))cdots )))) = 2^{2018} $$
What is the length of the shortest such sequence?

For example, one sequence that qualifies is
$$ underbrace{fcirc fcirc cdots circ f circ f}_{2018;ftext{s}} $$
But that is not the shortest possible sequence, because
$$ gcirc underbrace{fcirc fcirc cdots circ f circ f}_{1009;ftext{s}} $$
also works.

You start with $1$. Then you can double that to get $2$, double it again to get $4$, then again to get $8$ and once again to get $16$. It took four steps to get to $16=2^4$.

But I could also have started at 1, then doubled it to $2$, then squared it to $4$, then squared that to get $16$. This way it only took three steps to reach $16=2^4$.

Your problem asks you to get to $2^{2018}$ instead of $2^4$, but the idea is the same. Use these two operations to get there as fast as possible.

Hint: Write everything as powers of $2$, starting with $1=2^0$. That makes all the numbers that appear a lot easier to handle.

It is easy to see that at first we find a way to have $2$ so we multiply by $2$

Now it is obvius that raising to the square is much easier to have a big number
if we raise to the square we will have $2^2$
now we will do a string of operations by multiplying or raising to the square
$$1 - 2-2^2-2^3-2^6-2^7-2^{14}-2^{15}-2^{30}-2^{31}-2^{62}-2^{63}-2^{126}-2^{252}-2^{504}-2^{1008}-2^{1009}-2^{2018}$$
So the answer is $17$.