Peanuts Gallery

We first observe that any number can be written as a sum of powers of two. So we start with 1+2+4+8+16+32+64+128+256+512=1023.

If we add 1024 to this we will get 2047 which exceeds 2015, so we have to squeeze in the 2015-1023=992 somewhere in between.

We write 992 by starting with the maximum power of two lower than the number given, to meet the condition of minimum number of steps that is we go on subtracting the powers of two from 992 greedily. so we go like this:
992-512=480
480-256=224
224-128=96
96-64=32
which is a power of 2 itself so we need no further decomposition.

Now we see that we need to add to our original sequence the numbers, 32, 64, 128, 256, 512 although they are already present. Here we use the fact that we can pick the same number of peanuts as the previous step.

So we have our final sequence as 1+2+4+8+16+32+32+64+64+128+128+256+256+512+512=2015.

Lovely question..broke my head for solving this.

First half is same as previous solution.

We get 1023 in 10 steps.
Remaining 992 = 512 + 480.
480 = 160 * 3 = (128 + 32)3 = 256+ 128 + 64+ 32.

These makes remaining 5 steps.

Hence, the answer is : $\boxed{15}$