Xavier Aptitude Test 2021 Paper

Let the 6 digit number be _ _ _ $42$ _
It is divisible by $2,3,5,7,11,13$
Since the number is divisible by both $2$ and $5$ the last digit of the number must be $0$.
The number is also divisible by $7,11,$ and $13$.
Hence the number must also be divisible by $7\times 11\times 13=7\times 11\times 13=1001$.
A number which is a multiple of $1001$ is of the form $\mathrm{abcabc}$.
This is because $\mathrm{abc}\times \left(1001\right)=\mathrm{abc}\times \left(1000+1\right)=\mathrm{abc}000+\mathrm{abc}=\mathrm{abcabc}$.
Hence the number is $420420$.
The first digit is $4$.