Sum of the given digits = 4+5+3+0 =12
in option (a), sum of digits = 9+6 = 15
in option (b), sum of digits = 7+8 = 15
in option (c), sum of digits = 4+2 = 6
in option (d), sum of digits = 5+4 = 9
Now, for option (d), the overall sum of digits will be 12+9=21 which is not divisible by 9
So, this option can be eliminated.
Now among 453096, 453078 and 453042, all the numbers are divisible by both 6 and 9.
So, we need to find which number is divisible by 7
Now,$453096=45\times\ 10^4+30\times\ 10^2+96$
Now,$10^2$≡2(mod 7) and$10^4$≡4(mod 7)
So,453096(mod 7)≡($45\times\ 4+30\times\ 2+96$)(mod 7)≡336(mod 7)≡0(mod 7) as 336 is divisible by 7
So, 453096 is divisible by 7
We can do similar operation in case of453078 and 453042
453078(mod 7)≡($45\times\ 4+30\times\ 2+78$)(mod 7)≡318(mod 7)≡3(mod 7) so it is not divisible by 7
453042(mod 7)≡($45\times\ 4+30\times\ 2+42$)(mod 7)≡282(mod 7)≡2(mod 7) so it is not divisible by 7
So, option A is the correct answer
