Let "n" be a number
9 + 3n = 57
3n = 57 - 9
3n = 48
n = 48/3
n = 16
Any odd multiple of 3 would be the counterexample of the given conjecture.
That is 3(2n - 1), for any natural number n, will be divisible by 3 but not divisible by 6 as 2 is not a factor.
<u>Examples:</u>
3(3) = 9
3(5) = 15
3(7) = 21 are some examples.
Answer:
5
Step-by-step explanation:
If k is three than 15÷3 is 5