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=>

If \(n = 5k,\) then \(7n + 2 = 7(5k) + 2 = 5(7k) + 2\) is not a multiple of \(5.\)

If \(n = 5k+1,\) then \(7n + 2 = 7(5k+1) + 2 = 5(7k) + 7 + 2 = 5(7k) + 9 = 5(7k+1)+4\) is not a multiple of \(5.\)

If \(n = 5k+2,\) then \(7n + 2 = 7(5k+2) + 2 = 5(7k) + 14 + 2 = 5(7k) + 16 = 5(7k+3)+1\) is not a multiple of \(5.\)

If \(n = 5k+3,\) then \(7n + 2 = 7(5k+3) + 2 = 5(7k) + 21 + 2 = 5(7k) + 23 = 5(7k+4)+3\) is not a multiple of \(5.\)

If \(n = 5k+4,\) then \(7n + 2 = 7(5k+4) + 2 = 5(7k) + 28 + 2 = 5(7k) + 30 = 5(7k+6)\) is a multiple of \(5.\)

Thus, \(n\) has remainder \(4\) when it is divided by \(5.\)

The possible values of \(n\) are \(104, 109, …, 199.\)

The number of possible values of \(n\) is \(\frac{(199-104)}{5} + 1 = 20.\)

Therefore, B is the answer.
Answer: B
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7n+2 either ends with 0 or 5, so 7n either ends with 8 or 3.
n is between 100 and 200, so n can be: 1ab. a can be digits from 0~9. b can be either 4 (7*4)or 3(7*9)
So, 1*10*2 = 20 values
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