Which of the following is necessarily a divisor of the number $ntimes1000+2 times n$ where $nlt 500$?

A three-digit number 'n' (less than 500) is taken. A six-digit number is formed by writing the number 'n' as first three digits and the number '2n' as the last three. Which of the following is necessarily a divisor of the number so obtained?

After this there is actually 4 options to choose $6,37,167,501$. The solution specified in my module is goes like this: "If the original number is $ab$c, the new $6-$digit number will be $abc \times 1002$", after this it is pretty straight forward, but I don't understand how they are forming "$abc \times 1002$" as the concatenated number.

Answer

If you start with n=abc [as digits], then shifting it over by three digits with (abc)*1000 would give abc000. If you wanted abcabc, you could then add n, for (abc)*1000+(abc), or (abc) * 1001. But you want the last three digits to be 2n, so it's (abc)*1000+(abc)*2, or abc*1002.