When life gives you lemons

Your strategy is to fill every bag with a distinct power of two number of lemons. You will have distributed

$\underbrace {2^0+2^1+2^2+\ldots+2^8+2^9}_{\text{10 bags}} = \underbrace {1+2+4+\ldots+256+512}_{\text{10 bags}} = 1023$

lemons, so all lemons will be used.

When your friend asks you for a number of lemons, you have to express this number in binary and give them the bags that correspond to the right powers of 2.

So, it always works.

Put $2^{n-1}$ lemons in the $n$ th bag.

Then the binary representation of the number tells you which bags to hand over! 🙂

Put one lemon in the first bag, then put twice as many lemons that are in the current bag into the next bag and so forth.

To hand over $n$ lemons, check to see whether the number is odd, if so hand over the first bag. Repeatedly move to the next bag, halve the number requested throwing away any remainder, if the resultant is odd, hand over that bag.

Added: 2019 June 25 2325: The first paragraph describes putting ascending power of starting from $2^0\Rightarrow 1$ . The second paragraph describes a method on implicitly generating the binary form of the requested number in a manner that facilitates the selection of the proper bags (that is, the correct powers of 2) . See [Russian peasant multiplication](https://en.wikipedia.org/wiki/Ancient_Egyptian_multiplication#Russian_peasant_multiplication).

Well let's look at this in binary, shall we?

We see that 1023 is ten 1s in binary. This is because $1023 = 1024 - 1 = 2^{10} - 1$ .

Hence, $1023 = {(1111111111) }_{ 2 }$ .

And for all numbers less than 1023, we can write them as 1s and 0s, but the number of 1s is always less than ten (in base10, you can think of this as 9999 having a greater digit sum ( $9+9+9+9=36$ ) than al numbers lower than it) eg.

$328={ (0101001000) }_{ 2 }\quad (three\quad 1s)\\ \quad 94={ (0001011110) }_{ 2 }\quad (five\quad 1s)\\ 755={ (1011110011) }_{ 2 }\quad (seven\quad 1s)$

Remembering the good ol' place value expansion we learn in 3rd class, we can rewrite them as sums of the power of 2 (since binary is base2) multiplied with the respective digits. This can also be viewed as the digits telling us whether to add a particular power of two or not .

(base 10 analogy : $58294 =$ (5th digit from the right) $\times { 10 }^{ 5-1 }+$ (4th digit from the right) $\times { 10 }^{ 4-1 }+$ (3rd digit from the right) $\times { 10 }^{ 3-1 }+$ (2nd digit from the right) $\times { 10 }^{ 2-1 }+$ (1st digit from the right) $\times { 10 }^{ 1-1 }=5\times{10}^{4}+8\times{10}^{3}+2\times{10}^{2}+9\times{10}^{1}+4\times{10}^{0}$ )

$1023=1\times { 2 }^{ 9 }+1\times { 2 }^{ 8 }+1\times { 2 }^{ 7 }+\dots 1\times { 2 }^{ 2 }+1\times { 2 }^{ 1 }+1\times { 2 }^{ 0 }\\ 328=0\times { 2 }^{ 9 }+0\times { 2 }^{ 8 }+0\times { 2 }^{ 7 }+\dots 0\times { 2 }^{ 2 }+0\times { 2 }^{ 1 }+0\times { 2 }^{ 0 }\\ \quad 94=0\times { 2 }^{ 9 }+0\times { 2 }^{ 8 }+0\times { 2 }^{ 7 }+\dots 1\times { 2 }^{ 2 }+1\times { 2 }^{ 1 }+0\times { 2 }^{ 0 }\\ 755=1\times { 2 }^{ 9 }+0\times { 2 }^{ 8 }+1\times { 2 }^{ 7 }+\dots 0\times { 2 }^{ 2 }+1\times { 2 }^{ 1 }+1\times { 2 }^{ 0 }$

Well now view things in terms of the bags. Just like how we interpreted the binary digit expansion, we can say that the digits tell us whether to give a particular bag or not . And as we can see from the binary representation of 1023, we can fill all the bags, and as seen from the binary representation of a number less than 1023, we only have to give bags that can be filled by 1023 lemons.

Hence we can give any number of lemons, we just have to fill the bags with power of 2, starting from one.

$\blacksquare$

As others solutions say, the number of lemons in each bag is just a power of 2. But why?

Having n numbers, you have to sum them in all possible ways in order to obtain all the numbers up to the requested value (1023).

Basically, you have to sum the ways to pick one number, the ways of pick two numbers, ..., the ways of pick n numbers (all of them); picked numbers will be summed and, if they're well chosen, it won't be repetitions.

In combinatorics notation ${n \choose 1}+{n \choose 2}+...+{n \choose n} = \displaystyle \sum_{k=1}^n {n \choose k}$ .

Thanks to Tartaglia/Pascal's triangle we know that sum equals to $2^n-1$ . And, since $1023=2^{10}-1$ , the answer to the problem question is affermative .

This better be a good friend, showing up out of nowhere asking for lemons while Im being tested....

Let the Nth bag contain 2^N lemons. Write n in the base 2; whenever a 1 appears in the representation, hand over that bag. Ed Gray

Simple. Binary. Put 1 lemon in the first bag, 2 lemon in the second bag, and so on. When somebody ask you to give n amount of lemon, you figure out what the number is in binary, then grab the lemon from the appropriate bags.

Why was it specified for n to be a positive integer? If you have bags from 2^0 to 2^9 you will be able to fulfil requests for any integer n value, just add to the odd value below n and then add 2^0 (1).

10 bags and 1023 solutions. You can’t transfer lemons out of the bags... but you can the crate. So fill the bags with 0 lemons and then fill them up to the required number when asked. Remember... the first request may be for 1023 lemons... so putting any lemon other than perhaps 1 lemon in the first bag makes you unable to give out 1023 lemons if so required.