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Explanation:

Let the 6 digit number be _ _ _ 42_
It is divisible by 2,3,5,7,11,13

Hence, the number is a multiple of 2 × 3 × 5 × 7 × 11 × 13 = 30030

Now, 30030 × k = _ _ _ 4 2 _

Last digit would definitely be 0.

Second last digits which is 2 should be the last digit of 3 × k. Hence, last digit of k should be 4.
​​​​​​​∴ k = 4 or 14 or 24 and so on.

The only possible value of k from the one's given above satisying the conditions given is 14.

​​​​​​​∴ 30030 × 14 = 420420

​​​​​​​∴ The first digit is 4.

Hence, option (e).

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