Soda can contest problem

Question

The store's rule is that you can use 3 empty cans to exchange for 1 can of soda. Alex's class has 17 students and each bought a can of soda at the beginning of the party. At most, how many more cans of soda can Alex's class get by exchanging?

First of all, I'm not sure I understood the question correctly. I thought that all it's really asking for is how many full cans of soda you can get from 17 empty ones, so basically just 17/3 which is 5 remainder 2, so 5 soda cans.

I don't think it's that easy, so I probably interpreted it wrongly.

Answer 1

If I start with 5 cans, I can get 2 free ones.

First, I take 3 of my 5 empty cans and exchange them for 1. I then take that new can and drink the soda, and use it along with the remaining 2 cans to get another free one.

How does that extend to starting with 17?

Answer 2

Let's go step by step:

You have $17$ the biggest multiple of $3$ smaller than $17$ is $15$ so after the first "round" we'll have 7 cans left (remeber that we've exchanged $\frac{15}{3} = 5$ cans)

$$17-15+5=7$$

The biggest smallest multiple of $3$ smaller than $7$ is $6$, so after this "round" we'll have 3 cans left. (remember those $\frac{6}{3} = 2$ cans that we've exhcanged)

$$7-6+2 = 3$$

And at last we can change this $3$ cans for $1$ and if we add up all those cans we've remebered, we'll end up with:

$$5+2+1 = 8 \text{ additional cans}$$