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ATQ,
Divisibility by 6 ⇒ divisible by both 2 and 3.
Sum of digits = 4 + 9 + 6 + 7 + 8 + a = 34 + a
To be divisible by 3: (34 + a) must be divisible by 3
Also, a must be even (for divisibility by 2), and not 0 or 5.
Even options: 0, 2, 4, 6, 8
Try a = 2 → 34 + 2 = 36 ✅
So, a = 2
(2×a + 3) = 2×2 + 3 = 7
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