This is asking a definite value (gcf) that needs to be same at every case...
To find out gcf between two nos. we need to factorize each of them and take the all possible common factor/s between them.
Frm stem 1:
K is 1 greater than j.That means all the combinations will have even or odd consecutive positive nos.For example k and j could be 11,10 or 10,11...in any case the greatest common factor is going to be 1.
So sufficient.
Fromstem 2:
KJ/5 is integer doesn't restrict the greatest common factor to be one fixed no. i.e. 1.so insufficient.
Amit
greatest common factor
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let's say k is a multiple of 2.CITI29 wrote: 1)k =j+1
then the other multiples of 2 are 2, 4, 6, ... more than k, as well as 2, 4, 6, ... less than k.
so j, which is only 1 more than k, isn't one of those.
same goes if k is a multiple of 3, or 4, or ... any positive integer (except 1).
this reasoning proves that k and j don't share any common factors (other than the trivial 1), so their gcf is 1.
the above poster has stated, correctly, that one of k & j must be even and the other odd. however, that condition alone doesn't mean that the gcf must be 1: for instance, 9 is odd and 12 is even, but their gcf is most certainly not 1.
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here's another clever way of thinking of the gcf: if the 2 numbers were the numerator and denominator of a fraction, what's the biggest # you could pull out of both to reduce the fraction?
if you make fractions out of these number pairs, in which one number is exactly 1 greater than the other - 7/8, 10/11, 18/19, etc. - it should be pretty clear to you that none of the resulting fractions are going to reduce.
that means the gcf is 1.
sufficient.
all this means is that at least one of j and k is divisible by 5. could be one or both of them, and, moreover, you know nothing whatsoever about divisibility by any other factors. so the gcf could actually be absolutely any number you want.CITI29 wrote:2)jk is divisible by 5.
or, if you like easily accessible concrete #s:
j = 3, k = 5 --> jk = 15 is divisible by 5 --> gcf = 1
j = 10, k = 5 --> jk = 50 is divisible by 5 --> gcf = 5
insufficient
answer = b
Ron has been teaching various standardized tests for 20 years.
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
