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:
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
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $3$. -- Dollars and Cents
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $2$. Dollars and Cents