Henry Ernest Dudeney/Modern Puzzles/87 - Another Street Puzzle/Solution

by : $87$

 * Another Street Puzzle

Solution

 * $(1): \quad$ The man lives at no. $239$ in a street with $169$ odd-numbered houses.


 * $(2): \quad$ The man lives at no. $408$ in a street with $208$ even-numbered houses.

Proof

 * Odd-Numbered Side

Let there be $m$ houses in the street, where we are told $50 < m < 500$.

Let the man live at no. $n = 2 k + 1$.

We have that:

Hence this problem is equivalent to finding the answers to Pell's Equation:
 * $n^2 - 2 m^2 = -1$

From the solution to Pell's Equation, the solutions to this are:


 * ${p_r}^2 - 2 {q_r}^2 = \paren {-1}^r$ for $r = 1, 2, 3, \ldots$

where $\dfrac {p_r} {q_r}$ are the convergent of the Continued Fraction Expansion of $\sqrt 2$.

It is the odd integer values of $r$ that we need in order to make the equal to $-1$.

From Continued Fraction Expansion of Root 2:

Hence we have $n$ and $m$ as:


 * $\begin{array} {r|r} m & n \\ \hline

1 & 1 \\ 5 & 7 \\ 29 & 41 \\ 169 & 239 \\ 985 & 1393 \\ \end{array}$

Because there are between $20$ and $500$ houses in the street, we know the man lives at no. $239$ in a street with $169$ houses on the odd-number side.


 * Even-Numbered Side

Let there be $m$ houses in the street, where we are told $50 < m < 500$.

Let the man live at no. $n$.

Let us suppose all the house numbers were divided by $2$.

Then the man is living on a street where the houses go $1$, $2$, $3$, $\ldots$, $m$.

This is then the same problem as $85$ - The House Number in the same collection.

In this case, the man lives at no. $204$ in a street of $288$ houses.

So, when we multiply the house numbers by $2$ so they are back up to what they were, we find that the man lives at no. $408$.