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ATQ,
Divisible by 15 ⇒ divisible by 5 and 3
Divisible by 5: last digit n must be 0 or 5
Divisible by 3: sum of digits must be divisible by 3
Case 1: n = 0
Sum = 3 + 8 + 1 + 6 + m + 0 = m + 18
Divisible by 3 ⇒ m = 0, 3, 6, 9
Case 2: n = 5
Sum = 3 + 8 + 1 + 6 + m + 5 = m + 23
Divisible by 3 ⇒ m = 1, 4, 7
Total values of m: 4 + 3 = 7
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