Split 2018 Ⅱ

Ten summands are not enough, since we can form at most $2^{10}=1024$ distinct sums out of them. But 11 works: We can write $2018=995+512+256+128+64+32+16+8+4+2+1$

Any positive integer $n<1024$ can be written uniquely as a sum of powers of 2, up to 512, and any $n$ with $2018 \geq n\geq 1024$ can be written as $n=995+(\text{a sum of powers of 2 up to 512})$ . Thus the answer is $\boxed{11}$ .

In general, $n$ is the number of digits of $N$ in base 2.

A charming problem that I have never seen in this form before. Thank you!

Just think about binary numbers, a 11 bit binary number can represent 0 to 2047