Henry Ernest Dudeney/536 Puzzles & Curious Problems/1 - Concerning a Cheque/Solution
536 Puzzles & Curious Problems by Henry Ernest Dudeney: $1$
- 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:
\(\ds 200 x + 2 y\) | \(=\) | \(\ds 100 y + x - 5\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 199 x - 98 y\) | \(=\) | \(\ds -5\) | rearranging |
This is a linear Diophantine equation.
Using the Euclidean Algorithm:
\(\text {(1)}: \quad\) | \(\ds 199\) | \(=\) | \(\ds \paren {-2} \times \paren {-98} + 3\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds -98\) | \(=\) | \(\ds \paren {-33} \times 3 + 1\) |
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:
\(\ds 1\) | \(=\) | \(\ds -98 - \paren {\paren {-33} \times 3}\) | from $(2)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -98 + 33 \times 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -98 + 33 \times \paren {1 \times 199 - \paren {\paren {-2} \times \paren {-98} } }\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -98 + 33 \times \paren {199 - 2 \times 98}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 33 \times 199 + 67 \times \paren {-98}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -5\) | \(=\) | \(\ds -5 \times \paren {33 \times 199 + 67 \times \paren {-98} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds -165 \times 199 + \paren {-335} \times \paren {-98}\) |
and so:
\(\ds x_0\) | \(=\) | \(\ds -165\) | ||||||||||||
\(\ds y_0\) | \(=\) | \(\ds -335\) |
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:
\(\ds x_0\) | \(=\) | \(\ds -165\) | ||||||||||||
\(\ds y_0\) | \(=\) | \(\ds -335\) | ||||||||||||
\(\ds a\) | \(=\) | \(\ds 199\) | ||||||||||||
\(\ds b\) | \(=\) | \(\ds -98\) | ||||||||||||
\(\ds d\) | \(=\) | \(\ds 1\) |
\(\ds x\) | \(=\) | \(\ds -165 - 98 t\) | ||||||||||||
\(\ds y\) | \(=\) | \(\ds -335 - 199 t\) |
In order to make $x$ and $y$ minimally positive, we see that $t = -2$:
\(\ds x\) | \(=\) | \(\ds -165 - 98 \paren {-2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -165 + 196\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 31\) | ||||||||||||
\(\ds y\) | \(=\) | \(\ds -335 - 199 \paren {-2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -335 + 398\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 63\) |
Thus we arrive at:
- $x = 31$
and:
- $y = 63$
and the solution follows.
$\blacksquare$
Sources
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $1$. Concerning a Cheque