Placing which of the following two digits at the right end of 4530 makes the resultant six digit number divisible by 6,7 and 9?

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 of 453078 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