SOLUTION: I am having trouble with a word problem. Here it is:
What is the smallest counting number n so that the
square root of 2n is a multiple of 3?
I think the answer is 18 but I d
Algebra.Com
Question 190074: I am having trouble with a word problem. Here it is:
What is the smallest counting number n so that the
square root of 2n is a multiple of 3?
I think the answer is 18 but I do not know how to show all work.
I think the square root of 2n has to be divisible by 3 but it also
has to be divisible by 2. 6 would be the smallest divisible by 2 and 3.
Thanks. JCE
This did not come out of the book.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You are 100% on the right track. 
must be divisible by 3 because that is one of the conditions of the problem. But it also must be divisible by 2 because for to be a rational number 
must have an even number of factors of 2. has one factor of 2 that you can see, so there must be another one buried in the n. So since must have a factor of 2 and a factor of 3, must have at least 2 factors of 2 and at least 2 factors of 3.
John
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