Henry Ernest Dudeney/Modern Puzzles/3 - Dollars and Cents/Solution

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Modern Puzzles by Henry Ernest Dudeney: $3$

Dollars and Cents
An American correspondent tells me that a man went into a store and spent one-half of the money that was in his pocket.
When he came out he found that he had just as many cents as he had dollars when he went in
and half as many dollars as he had cents when he went in.
How much money did he have on him when he entered?


Solution

The total amount he had on him was $\$ 99.98$.


Proof

Let $S$ be the amount he started with: $S_d$ dollars and $S_c$ cents.

Let $F$ be the amount he finished with: $F_d$ dollars and $F_c$ cents.


We recall the conversion factors:

$100$ cents make one dollar.

Hence any cent quantities in either $S$ or $F$ cannot be greater than $99$.

That is:

$S_c < 99$
$F_c < 99$


We are given that:

\(\ds S\) \(=\) \(\ds 2 F\) ... spent one-half of the money that was in his pocket.
\(\ds F_c\) \(=\) \(\ds S_d\) When he came out he found that he had just as many cents as he had dollars when he went in
\(\ds 2 F_d\) \(=\) \(\ds S_c\) and half as many dollars as he had cents when he went in.

We have that:

\(\ds S\) \(=\) \(\ds 100 S_d + S_c\) where $S$ cents is the money he started out with
\(\ds F\) \(=\) \(\ds 100 F_d + F_c\) where $F$ cents is the money he came home with
\(\ds \) \(=\) \(\ds 100 \dfrac {S_c} 2 + S_d\) as $2 F_d = S_c$
\(\ds \) \(=\) \(\ds 50 S_c + S_d\)
\(\ds \leadsto \ \ \) \(\ds 100 S_d + S_c\) \(=\) \(\ds 2 \paren {50 S_c + S_d}\) as $2 F = S$
\(\ds \leadsto \ \ \) \(\ds 99 S_d\) \(=\) \(\ds 98 S_c\) simplifying


The smallest values of $S_d$ and $S_c$ that satisfy the above equation are:

\(\ds S_d\) \(=\) \(\ds 99\)
\(\ds S_c\) \(=\) \(\ds 98\)

As $S_c \le 99$ it follows that there can be no other solution.

Hence:

$D = \$ 99.98$

$\blacksquare$


Sources

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