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

by : $3$

 * Dollars and Cents

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:

We have that:

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

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

Hence:


 * $D = \$ 99.98$