Fascinating Offer for Greedy Kiddo

To start with, the kid buy's $1729$ chocolates with $Rs.1729$ . And hence he will now have $1729$ wrappers.

According to the question the kid can buy $1$ more chocolate with every $3$ wrappers, so he can now buy $\dfrac{1729}{3}$ chocolates.

But it leaves $1$ as the remainder, so the kid can buy $576$ chocolates.

Similarly, the remaining $576$ wrappers, will get him $\dfrac{576}{3}=192$ chocolates.

Moving on, he buys $64$ chocolates, which will get him $\dfrac{64}{3}=21$ chocolates, with a remainder of $1$ wrapper.

Next, he can buy $\dfrac{21}{7}=7$ chocolates with the $21$ wrappers. Now, he uses the remaining $2$ wrappers and gets $\dfrac{9}{3}=3$ more chocolates.

At last, the Greedy Kiddo uses the last $3$ wrappers to get $1$ more chocolate.

So the maximum number of chocolates he can buy $= 1729+576+192+64+21+7+3+1=\boxed{2593}$

P.S: Please don't eat those many chocolates it can cause CAVITIES!

The answer is that with the given restrictions and the given exchange offer one can get 2593 sweets if one spends 1729 Rupees.

I was interested in generalizing the problem to any integral amount of Rupees $\geq 1$ to start with.

For the first couple of integral Rupees we can easily figure out the maximum number of sweets that can be "bought" by hand and then seek the pattern. The first few ordered pairs (Rupees , Sweets) are:

$(1,1), (2,2), (3,4), (4,5), (5,7), (6,8), (7,10), (8,11), (9,13), (10,14)$

The pattern is easily figured out and understood. The increase in the number of sweets is alternating 1 and 2 as the amount of Rupees steadily increases with 1.

If we only look at even Rupees amounts the sequence starts with 2 and increases with 1 + 2 = 3. As this series is a arithmetic sequence we therefore denote:

$N = \lfloor \frac{ \text{Rupees} }{2} \rfloor \qquad \text{ with } \qquad R = \text{Rupees} - 2 \cdot N$

This basically means that R is the remainder after division by 2.

Now the maximum number of sweets that can be bought with any integral amount of Rupees $\geq 1$ is:

$\text{Sweets} = 3 \cdot (N - 1) + 2 \cdot (R + 1)$

So if one spends 1729 Rupees on sweets then $N = \lfloor \frac{1729}{2} \rfloor = 864$ , $R = 1729 - 2 \cdot 864 = 1$ and therefore the maximum number of sweets that can be "bought" is:

$\text{Sweets} = 3 \cdot (864 - 1) + 2 \cdot (1 + 1) = 3 \cdot 863 + 2 \cdot 2 = 2593$

1729 Rs. = 1729 toffees

1729 toffees = 1729 wrappers = 576 toffees + 1 ex1 wrapper

576 toffees = 576 wrappers = 192 toffees

192 toffees = 192 wrappers = 64 toffees

64 toffees = 64 wrappers = 21 toffees + 1 ex2 wrapper

21 toffees = 21 wrappers = 7 toffees

7 toffees = 7 wrappers = 2 penultimate toffees + 1 ex3 wrapper

1 ex1 wrapper + 1 ex2 wrapper + 1 ex3 wrapper = 3 wrappers = 1 ex toffee

1 ex toffee + 2 penultimate toffees = 3 toffees

3 toffees = 3 wrappers = 1 last toffee

Total toffees (sum of all the toffees in bold):

1729 + 576 + 192 + 64 + 21 + 7 + 3 + 1 = 2593 toffees