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divisors, integers

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divisors, integers

by mberkowitz » Wed Oct 01, 2008 2:55 pm
Is n/18 an integer?

1.5n/18 is an integer

2.3n/18 is an integer

OA is C

This question was posted and answered some time ago. Might one of the wizards break this down?

Thanks very much.
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by nirupshetty » Wed Oct 01, 2008 5:25 pm
1) the 1'st eqn is satisfied for n = 18/5.. how the question doesn't come up to be an integer.. i.e. NOT SUFF
2) this one is more simple to analyze.. for n = 6 the question doesnt hold true.. i.e. NOT SUFF

however, together only 90 satisfies the condition. therefore, C
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Re: divisors, integers

by Morgoth » Wed Oct 01, 2008 8:05 pm
mberkowitz wrote:Is n/18 an integer?

1.5n/18 is an integer

2.3n/18 is an integer

OA is C

This question was posted and answered some time ago. Might one of the wizards break this down?

Thanks very much.
Here is clear method:

statement (1)
5n/18 = integer

let n = 18/5
5*18/5 /18 = 18/18 = 1 (integer)
n=18/5 not integer, Insufficient.

statement (2)
3n/18 = integer

n/6 = integer
n= 6
6/6 =1 integer
6/18 = 1/3 not integer, Insufficient.

combining (1) & (2)
5n/18 & 3n/18
any integer for n which satisfies above two conditions will also satisfy the condition for n/18.
Sufficient.

Thus, C.
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by mberkowitz » Wed Oct 01, 2008 8:20 pm
i still dont understand why c works. granted i see why 1 and 2 are insufficient, because there exist values for which it works in the statement and not in the root, but how do you assume c? in which case would the answer be e? also if i asked you to find the lowest possible value of n given all info how would you find it, intelligible for the average beer drinker?
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by mberkowitz » Wed Oct 01, 2008 8:21 pm
to get c, are you essentially plugging in values for both and finding that any value that works for both statement will also work for the root? how do you come up with that value?
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