Tough Problem

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Tough Problem

by prachi18oct » Mon Jun 15, 2015 7:33 am
If m and q are divisors of c, is mq a divisor of c that is not c or 1?

(1) mq < c
(2) m and q have no common divisors.

A)Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
B)Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
C)BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D)EACH statement ALONE is sufficient.
E)Statements (1) and (2) TOGETHER are NOT sufficient.


Lets take numbers c = 12, m , q can be from (1,2,3,4,6,12)

1) mq < c => m,q cannot be 2,6/3,4/4,6/6,12
m q can be 1,2;1,3;1,4;1,6;2,3 or 2,4

From all the possible combinations,
if m,q = 2,3 then it is divisor of 12
if m,q = 2,4 then it is not divisor of 12
SO INSUFFICIENT

2)m , q no common divisors

so we can eliminate some sets from previous selection.
m,q can be 3,4; 1,2; 1,3 ;4,1; 6,1; 2,3
All of them have mq as common divisor of c but if m,q = 3,4 then mq = 12 = c. SO NO
INSUFFICIENT.

Together,
we have only 1,2; 1,3 ;4,1; 6,1; 2,3 . ALl of these combinations satisfy mq < c and mq divisor of c.
HENCE SUFFICIENT.


Please let me know if I missed something or assumed anything.
Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Mon Jun 15, 2015 7:53 am
prachi18oct wrote:If m and q are divisors of c, is mq a divisor of c that is not c or 1?

(1) mq < c
(2) m and q have no common divisors.
Are you sure you transcribed the question properly.
To begin, most (all maybe) GMAT divisibility questions restrict the values to POSITIVE integers, and this question does not do that.

Also, m and q have no common divisors is hard to swallow. If m and q are non-zero integers, they will at least have 1 as a common divisor. They will also have -1 as a common divisor (if we're not restricting the integers to POSITIVE values.

What's the source of this question?

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by prachi18oct » Mon Jun 15, 2015 7:56 am
Hi Brent,

This is from bellcurves practice question bank.

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by nikhilgmat31 » Thu Jun 18, 2015 3:58 am
really difficult question, Brent.

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by nikhilgmat31 » Fri Jun 26, 2015 3:18 am
1 is common divisor of m & q

so why we took 1,2 or 1,3 or 1,6 is statement 2

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by Matt@VeritasPrep » Mon Jun 29, 2015 5:48 pm
Since most math tests/number theory books/etc. assume that a divisor is positive unless otherwise specified, I'll make that assumption here.

The prompt tells us that m and q are divisors of c, so by our assumption 1 ≤ m ≤ c and 1 ≤ q ≤ c.

S1:: mq < c

Trying numbers seems easiest. Suppose c = 6, m = 2, and q = 2. (m and q could certainly be the same divisor!) In this case mq < c, but mq is NOT a factor of c.

But we could also have c = 12, m = 2, and q = 3. In that case mq < c AND mq is a factor of c. So S1 is NOT sufficient, as we get conflicting results.

S2:: m and q have no common divisors

This is impossible, since 1 is a divisor of both m and q ... and the question is broken. But let's assume that the question meant to say m and q have no common divisors other than 1 itself (i.e. no common prime divisors).

In this case, we could have c = 15, m = 3, q = 5, in which mq is a divisor of c but IS equal to c itself. We could also have c = 30, m = 3, q = 5, in which case mq is a divisor of c AND 1 < mq < c. S2 is NOT sufficient.

S1 + S2::

We could have m = 1, q = 1, and mq = 1, in which case the answer is NO, or we could have the examples above in which the answer is YES.

However, I'm assuming that the question writer also wanted to exclude the case in which m = q = 1 ... but who knows. (They probably said the answer is C, but S2 is so tricky for me to fix that it's hard to know what other assumptions they were making when they wrote this.)

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by nikhilgmat31 » Tue Jun 30, 2015 4:30 am
Please explain statement 2

2)m , q no common divisors

so we can eliminate some sets from previous selection.
m,q can be 3,4; 1,2; 1,3 ;4,1; 6,1; 2,3
All of them have mq as common divisor of c but if m,q = 3,4 then mq = 12 = c. SO NO
INSUFFICIENT.