Joseph takes any three-digit number, such as 347, and writes it twice to make a six-digit number, 347347. By which of the following is every such six-digit number always divisible?

Explanation

1. 01

   Writing abc twice gives abc × 1,000 + abc = abc × 1,001.

2. 02

   $1{,}001 = 7 \times 11 \times 13$.

3. 03

   So every such number is divisible by 7, 11 and 13; of the choices, 13.