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ATQ,
48 = 2 × 3 × 8 So, 102a32 is divisible by both 3 and 8 Since 102a32 is divisible by 3, so sum of the digits must be divisible by 3 So, (1 + 0 + 2 + x + 3 + 2) = (a + 8) must be divisible by 3 So, (a + 8) = 9 or 12 or 15 So, a = 1 or 4 or 7 For 102a32 to be divisible by 8, a32 must be divisible by 8 So, a = 2 or 4 or 6 or 8 Since the number is divisible by both 3 and 8, So a = 4
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