Henry Ernest Dudeney/536 Puzzles & Curious Problems/1 - Concerning a Cheque/Solution

by : $1$

 * Concerning a Cheque
 * A man went into a bank to cash a check.
 * In handing over the money the cashier, by mistake, gave him dollars for cents and cents for dollars.
 * He pocketed the money without examining it, and spent a nickel on his way home.
 * He then found that he possessed exactly twice the amount of the check.
 * He had no money in his pocket before going to the bank.
 * What was the exact amount of that check?

Solution
The amount on the the check was $\$ 31.63$.

After leaving the bank, he had $\$ 63.31$.

After arriving home, he had $\$ 63.26$, which is twice $\$ 31.63$.

Proof
Let $C$ be the amount written on the check.

Let $C$ consist of $x$ dollars and $y$ cents.

Let $C_1$ be the value of the check in cents.

Let $C_2$ be the money the man left the bank with in cents.

Let $C_3$ be the money the man arrived home with in cents.

We have:
 * $C_1 = 100 x + y$

After coming out of the bank, the man has:
 * $C_2 = 100 y + x$

After arriving home, the man has:
 * $C_3 = 100 y + x - 5$

But we have:
 * $C_1 \times 2 = C_3$

which leads us to:

This is a linear Diophantine equation.

Using the Euclidean Algorithm:

Thus we have that:
 * $\gcd \set {199, -98} = 1$

which is a divisor of $-5$:
 * $-5 = -5 \times 1$

So, from Solution of Linear Diophantine Equation, a solution exists.

Next we find a single solution to $199 x - 98 y = -5$.

Again with the Euclidean Algorithm:

and so:

is a solution.

From Solution of Linear Diophantine Equation, the general solution is:


 * $\forall t \in \Z: x = x_0 + \dfrac b d t, y = y_0 - \dfrac a d t$

where $d = \gcd \set {a, b}$.

In this case:

In order to make $x$ and $y$ minimally positive, we see that $t = -2$:

Thus we arrive at:


 * $x = 31$

and:
 * $y = 63$

and the solution follows.