Christmas problem, the salesman with the nuts

Question

At the Christmas market, a man was selling nuts in a market stall.

• The first person bought one nut, the next customer bought two nuts, the next bought four, and so on.

That is, every new customer acquired twice as many nuts as the previous one. Last customer of the day bought 50 kg of nuts, and after that the salesman had only one nut left. How much weighed all the nuts in the beginning of the day? (All nuts are the same.)

Answer 1

Well we can say that the number of nuts sold, $N$, is equal to $$N=\sum_{k=0}^{c}2^k$$ where $c$ is the number of customers. Then if the last customer bought $50$kg of nuts that means he bought $$N_f=\frac{50}{m}$$ nuts where $m$ is the mass of the nuts. That means that $$2^c=\frac{50}{m}$$ therefore $$c=\log_{2}\frac{50}{m}$$ So in total, the mass of all the nuts in the begging was $$m_i=m(\sum_{k=0}^{\log_{2}\frac{50}{m}}2^k+1)$$ This is good but we can take it a step further, we know that since $2^c=\frac{50}{m}$, that $\frac{50}{m}$ must only have prime factors of $2$. To do this since $50=2(5^2)$ we know $m$ must be of the form $$m=\frac{25}{2^j}$$ where $j$ is an integer. Therefore $$2^c=2*2^j$$ $$c=j+1$$ This means we can now say that $$m_i=\frac{25}{2^j}\left(\sum_{k=0}^{j+1}2^k+1\right)$$ Which simplifies down to $$m_i=\frac{25}{2^j}(2^{j+2}-1+1)$$ So we conclude $$m_i=100kg$$

Answer 2

The last customer being the $n-th$ customer, they purchased $2^n$ nuts. As he number of nuts in total is simply twice this, the weight ( on average ) is also twice this. Its 100kg.

The progression you describe is simply powers of two, and adding successive powers of two gets you :

$$\sum_0^n{2^k} = 2^{(n+1)}-1$$

As there is one nut left over we simply had a total of $2^{n+1}$ nuts starting, which is twice as many as the last purchaser.

Answer 3

Well, ok, there were only two nuts, and after the first was sold that was it. So the first (and last) customer bought a single $50$ kg nut, which together with the remaining nut makes a total of $100$ kg of nuts.

Hmm, maybe there were enough nuts for two customers. The first bought a nut, the second bought two nuts and so two nuts weigh $50$ kg, this makes $25$ kg each nut, and total $4$ nuts, $100$ kg.

Looks like there is a pattern ...

Answer 4

Your question on how much weighed all the nuts in the beginning of the day is simple, the answer is:

$50 + \dfrac {50}{2} + \dfrac {50}{2^2} + ... + \dfrac {50}{2^{n_0}} + \dfrac {50}{2^{n_0}}=100$, so we see that the answer is independent of the number of customers.

But if you asked what is the number of customers then the answer would be that it is not uniquely determined from the data that we have in the statement of the problem, in fact it seems that the number of customers can be any natural number $\geq 2$.

Answer 5

The zeroth customer put the last nut on layaway (he was really poor). Take him into account, and each customer bought exactly as much as had all the customers prior put together.

So the total inventory at the start was twice what the last customer bought, or $100$ kg.