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If (3N + 5) represents a multiple of 10, Then what is true about "N"

A. N is even
B. N is Odd
C. N is a factor of 30
D. N is a factor of 50

User Raz Ronen
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1 Answer

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Answer: B: N is odd

Explanation:

If 3N + 5 is a multiple of 10, then it can be written as:

3N + 5 = 10k, where k is an integer.

Subtracting 5 from both sides, we get:

3N = 10k - 5

Dividing both sides by 3, we get:

N = (10k - 5) / 3

For N to be an integer, 10k - 5 must be divisible by 3.

We know that any multiple of 10 that is divisible by 3 must have a units digit of 0, and the sum of its digits must be divisible by 3.

The units digit of 3N + 5 is 5, which is not 0, so N is not even. Therefore, option A is not correct.

The sum of the digits of 3N + 5 is 3 + 5 = 8, which is not divisible by 3. Therefore, N is not a factor of 30. Therefore, option C is not correct.

Similarly, since 3N + 5 is not divisible by 5, N cannot be a factor of 50. Therefore, option D is not correct.

Therefore, the only possible option is B: N is odd.

To see why, consider the following: if N is odd, then 3N is odd, and 3N + 5 is even. If 3N + 5 is even, then it must be divisible by 2, which means it is also divisible by 10 (since it is given that 3N + 5 is a multiple of 10). Therefore, N must be odd.

User Alexey Orlenko
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