Hi Brent,

This is from bellcurves practice question bank.

really difficult question, Brent.

1 is common divisor of m & q

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

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.)

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.