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A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of $5$ children. How many toffees were there in his stock initially?
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Let $x$ be the number of toffees in the stock initially.

$$\begin{array}{} & \text{Gives} & \text{Left with} \\ 1^{\text{st}} & \frac{x}{2} + 1 & \frac{x}{2}-1 \\ 2^{\text{nd}} & \frac{1}{2}\left( \frac{x}{2} – \right) + 1 & \frac{1}{2}\left(\frac{x}{2} – 1\right) – 1 \\ 3^{\text{rd}} & \frac{1}{2}\left[ \frac{1}{2}\left(\frac{x}{2} – 1\right) – 1 \right] + 1 & \frac{1}{2}\left[ \frac{1}{2}\left(\frac{x}{2} – 1\right) – 1 \right]-1 \\ \vdots & \vdots \quad \vdots \quad \vdots & \vdots \quad \vdots \quad \vdots \\ 5^{\text{th}} &  \vdots \quad \vdots \quad \vdots & 0  \end{array}$$
His total stock exhausts after he takes the case of $5$ children.

So, $ \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{2}x-1 \right) – 1 \right) – 1 \right) – 1 \right) – 1 = 0$

$ \Rightarrow \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{4}x – \frac{1}{2} – 1 \right) – 1 \right) – 1 \right) – 1 = 0$

$ \Rightarrow  \frac{1}{2} \left( \frac{1}{2} \left( \frac{1}{8}x – \frac{1}{4} – \frac{1}{2} – 1 \right) – 1 \right) – 1 = 0$

$ \Rightarrow \frac{1}{2} \left(\frac{1}{16}x – \frac{1}{8} – \frac{1}{4} – \frac{1}{2} – 1 \right) – 1= 0$

$ \Rightarrow \frac{1}{32}x – \frac{1}{16} – \frac{1}{8} – \frac{1}{4} – \frac{1}{2} – 1= 0$

$ \Rightarrow \frac{x – 2 – 4 – 8 – 16 – 32}{32} = 0$

$ \Rightarrow x – 62 = 0$

$ \Rightarrow \boxed{x = 62}$

$\therefore 62$ toffees were there in his stock initially.


$\textbf{Second Method:}$

$$\begin{array}{|c|c|c|c|} \hline \text{Toffees} & \text{Children} & \text{Gives} & \text{Left with} \\\hline  a\;(\text{Initially}) & 1^{\text{st}} & \frac{a}{2} + 1 & \frac{a}{2}-1 \\  b & 2^{\text{nd}} & \frac{b}{2} + 1 & \frac{b}{2} – 1 \\ c & 3^{\text{rd}} & \frac{c}{2} + 1 & \frac{c}{2} – 1 \\  d & 4^{\text{th}} & \frac{d}{2} + 1 & \frac{d}{2} – 1 \\  e & 5^{\text{th}} & \frac{e}{2} + 1 & \underbrace{\frac{e}{2} – 1}_{0} \\\hline  \end{array} $$
Here, $\frac{e}{2} – 1 = 0$

$ \Rightarrow \frac{e}{2} = 1$

$ \Rightarrow \boxed{e=2}$

And, $\frac{d}{2} – 1 = e$

$ \Rightarrow \frac{d}{2} - 1 = 2$

$ \Rightarrow \frac{d}{2} = 3$

$ \Rightarrow \boxed{d = 6}$

Similarly $\frac{c}{2} – 1 = d$

$ \Rightarrow \frac{c}{2} - 1 = 6$

$ \Rightarrow \frac{c}{2} = 7$

$ \Rightarrow \boxed{c = 14}$

And, $\frac{b}{2} – 1 = c$

$ \Rightarrow \frac{b}{2} - 1 = 14$

$ \Rightarrow \frac{b}{2} = 15$

$ \Rightarrow \boxed{b = 30}$

And, $\frac{a}{2} – 1 = b$

$ \Rightarrow \frac{a}{2} - 1 = 30$

$ \Rightarrow \frac{a}{2} = 31$

$ \Rightarrow \boxed{a = 62}$

Correct Answer$: 62$

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