The integers k and n are such that 4 < k < n and k is

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258. The integers k and n are such that 4 < k < n and k is not a factor of n. If R is the remainder when n is divided by k, is R > 2?
1. Thre greatest common factor of k and n is 4.
2. The least common multiple of k and n is 84.
Source: — Data Sufficiency |

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by puneetkhurana2000 » Sun Dec 16, 2012 4:13 pm
Statement 1) The greatest common factor of k and n is 4.

Assuming some values we get
k = 8 and n = 12, remainder = 4
k = 8 and n = 16, remainder = 0

Not Sufficient!!!

Statement 2) The least common multiple of k and n is 84.

Assuming some values we get
k = 7 and n = 12, remainder = 5
k = 12 and n = 28, remainder = 4
k = 28 and n = 42, remainder = 14

Remainder is always > 2.

Sufficient!!!

Answer B.

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by puneetkhurana2000 » Sun Dec 16, 2012 4:21 pm
Can someone explain this in detail?

As both statements together we get, 4*a * 4*b = 84 * 4, solving we get a*b = 21 and only values that satisfy this keeping 4<k<n in mind are k = 4*3 = 12 and n = 4*7 = 28.

If we divide n by k we get a remainder as 4, which is greater than 2. Hence we are certain.

But can we answer the question with such certainty by taking into account only Statement 2(shown in my previous reply to this thread).

Thanks

Puneet

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by tranhang233 » Sun Dec 16, 2012 10:51 pm
I think the correct answer is D.
in your explanation:
+ Statement 1: The greatest common factor of k and n is 4.
Assuming some values we get
k = 8 and n = 12, remainder = 4
k = 8 and n = 16, remainder = 0 (8 is the factor of 16, but the stem is that k is not a factor of n)
--> statement 1 is sufficient
--> correct answer is D

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by puneetkhurana2000 » Sun Dec 16, 2012 11:32 pm
Thanks Tranhang, I missed that and Answer should be D.

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by The Iceman » Mon Dec 17, 2012 9:22 am
Statement I:

If gcd(n,k)=4, let n=4a, k=4b

We have 4a = 4bx + R => R must be a multiple of 4. Hence, Statement 1 is sufficient.

Statement II:

LCM(n,k)= 84

If(n,k)= (12,7) => R=5

If(n,k)= (14,12) => R=2

So, statement II is not sufficient.

Hence, choice A is correct