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

We need to place two digits at the right end of $4530$ to make a six-digit number divisible by $6$, $7$, and $9$.

The number looks like: $4530\_ \_$

If a number is divisible by both $6$ and $9$, it must be divisible by $2$, $3$, and $9$. Since divisibility by $9$ already covers divisibility by $3$, we really need the number to be divisible by $2$, $9$, and $7$.

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Let's check option (a): $453096$

Divisibility by $2$:

Last digit $= 6$ (even) ✅

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Divisibility by $9$:

A number is divisible by $9$ if the sum of its digits is divisible by $9$.

$4 + 5 + 3 + 0 + 9 + 6 = 27$

$\dfrac{27}{9} = 3$

✅

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Divisibility by $6$:

Since it's divisible by both $2$ and $3$ (since $27$ is divisible by $3$) ✅

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Divisibility by $7$:

$\dfrac{453096}{7} = 64728$

Exact division, no remainder ✅

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$453096$ is divisible by $6$, $7$, and $9$ — all conditions satisfied.

You can quickly eliminate other options by checking the easiest rules first:

• If the last digit is odd → not divisible by $2$ → not divisible by $6$ → eliminated immediately
• If the digit sum is not a multiple of $9$ → not divisible by $9$ → eliminated

Always check divisibility by $2$ and $9$ first — they're the fastest. Only if both pass, then check divisibility by $7$.

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The two digits are $9$ and $6$, making the number $453096$.