Trying to get 2018

Logic Level 3

I have five kinds of cards(there are unlimited amount of them), $\large\boxed{ 1 }$ , $\large\boxed{ + }$ , $\large\boxed{\times}$ , $\large\boxed{(}$ , $\large\boxed{)}$ .

I can get numbers by putting the cards in a row (no exponential),and I don't want to put two $\large\boxed{ 1 }$ s together.

For example,I can get to $14$ by $(1+1)((1+1)(1+1+1)+1)$ but not $11+1+1+1$ .

I want to use these cards to get to $2018$ ,what is the least amount of $\large\boxed{ 1 }$ s I have to use?

See similar problem here

The answer is 23.

1 solution

X X
May 13, 2018

$2018=2(1008+1)=2(2^4\times3^2\times7+1)$ $=(1+1) [ (1+1)(1+1)(1+1)(1+1)(1+1+1)(1+1+1){\color{#3D99F6}((1+1)(1+1+1)+1)}+1 ]$

Trying to get 2018

The answer is 23.

1 solution

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