Henry Ernest Dudeney/Modern Puzzles/55 - The Repeated Quartette/Solution

by : $55$

 * The Repeated Quartette

Solution

 * $273863$

Proof
The key point is that a number of the form $\sqbrk {abcdabcd}_{10}$ is equal to $10001 \times \sqbrk {abcd}_{10}$.

We also have that:


 * $10001 = 73 \times 137$

and:
 * $365 = 5 \times 73$

So the $8$ digit number we end up with is a multiple of $5 \times 73 \times 137 = 50005$.

All we need to understand is that it has $2$ repeated sets of $4$ digits and ends in $5$.

The largest one of those is:
 * $99959995 = 273863 \times 365$

and the job is done.