Repeaters
Problem
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594).
Whatever digits you choose the number will always be divisible by 7 and by 11 and by 13, in each case without a remainder.
Can you explain why?
Student Solutions
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 $13$.