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## 'Repeaters' printed from http://nrich.maths.org/

Tim from Gravesend Grammar School and Mohammad Afzaal Butt both sent us similar solutions to the problem. Well done Tim and Mohammad. Here is Mohammad's solution:

Let the three digit number be $xyz$. Hence the six digit number will be $xyzxyz$. Now

$$\eqalign { xyzxyz &= 100000x + 10000y + 1000z + 100x + 10y + z \cr

&= 100100x + 10010y + 1001z \cr

&= 1001 (100x + 10y + z) \cr

&= 7 \times 11 \times 13 (100x + 10y + z)} $$ Hence the number $xyzxyz$ is always divisible by $7$, $11$ and $13$.