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KAPLAN NUMBER PORPERTIES

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KAPLAN NUMBER PORPERTIES

by resilient » Tue Apr 08, 2008 12:22 am
If 6n is an integer, whic of the following must be an integer?

a.n
b.2n
c.4n
d.9n
e.18n


qa is E

I chose n =1/2 , 2 etc but cant rationalize the answer
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by sandeepdeb » Tue Apr 08, 2008 12:50 am
I think the answer should be E. 6n is an interger means that n is a rational number of the form p/q where p and q are intergers and are prime to each other. q can be anything of 1,2,3 or 6. If we take 18n, then 18 is divisible by all the four possible values of q. So, while the others can also be integers, they can also not be integers based on certain values of q. 18n is surely an integer in that respect.

Hope I was able to clarify myself.
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by codesnooker » Tue Apr 08, 2008 2:09 am
Enginpasa1, its very simple.

Given statement: 6n is an integer.

For this kind of equation, if you are trying to solve the question by plugging values, then always think n = (1/n's multiple).

i.e for our case n = 1/6 (because, here 6 could be maximum value in the denominator that can still make "6n" as integer. i.e. for general case the denominator's maximum value could be only multiple of n.

Now try with all the given choices. n, 2n, 4n and 9n all fails to satisfy to make 6n as an integer.

If you are looking for mathematical approach then sandeep has explained it well.

Goodluck!!!
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by richardwang6430 » Tue Apr 15, 2008 6:10 pm
Obviously, it is E. 18n/6n=3
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