f2001290 wrote:k, m, and t are positive integers and k/6 + m/4 = 12t, do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

We have 2 k + 3 m = 144 t, where k, m, and t are positive integers. The question otherwise is, "Are t and 12 not co prime?"

(1) If k is a multiple of 3, so is 144 or 12; hence, t may or may not have 3 as its factor. Insufficient.
(2) If m is a multiple of 3, 3 m is a multiple of 9, so is 144; hence, t may or may not have 3 as its factor. Insufficient.

When taken together, let's take k = 3 a and m = 3 b where a and b are positive integers, so that 6 a + 9 b = 144 t, or 2 a + 3 b = 48 t. Now, if a is a multiple of 3, t may or may not have 3 as its factor, otherwise t and 12 must have a common factor greater than 1.

[spoiler]E[/spoiler]

My stab at this one is a bit more convoluted:

basic equation 2k+3m = 144t

stmt 1 let k = 3p, 2p + m = 48t can we find p,m and t such that the equation is satisfied when t is a factor of 12 and not a factor. For t = 4, m = 100, p = 48 and t = 5, m = 120, p = 60, we see that t can share a factor with 12 and not also. hence insufficient.
stmt 2 let m = 3q, 2k + 9q = 144t can we find k,q and t such that the equation is satisfied when t is a factor of 12 and not a factor. For t = 2, q = 10, p = 96 and t = 5, p = 40, k = 180, we see that t can share a factor with 12 and not also. hence insufficient.
Stmt 1 and 2, 2p + 3q = 48t as above we can show that t can have a factor common with 12 and not and still satisfy the equation. Hence, the choice is E

I believe Answer should be A.

What is the source of this question?

kvcpk wrote:I believe Answer should be A.

What is the source of this question?

please go ahead

k/6 + m/4 = 12t

12 and t should/shouldnot have a common factor greater than 1.
Which means that t and 12 should be relatively prime.
implies -> t should not have any multiples of 2 or 3.

(2k + 3m)/12 = 12t
(2k + 3m)/144 = t

given that k,m,t are positive integers.

So, (2k+3m)/144 should be an integer.

given that k is multiple of 3. so let k=3x
(2(3x)+3m)/144 = t
taking 3 out,
3 * (some integer) = t

so t and 12 have a commn factor of 3. So Chose A.

But I am sure that there will be a mistake in my procedure.

kvcpk wrote:k/6 + m/4 = 12t

12 and t should/shouldnot have a common factor greater than 1.
Which means that t and 12 should be relatively prime.
implies -> t should not have any multiples of 2 or 3.

(2k + 3m)/12 = 12t
(2k + 3m)/144 = t

given that k,m,t are positive integers.

So, (2k+3m)/144 should be an integer.

given that k is multiple of 3. so let k=3x
(2(3x)+3m)/144 = t
taking 3 out,
3 * (some integer) = t

so t and 12 have a commn factor of 3. So Chose A.

But I am sure that there will be a mistake in my procedure.

I think the part I highlighted in red is your mistake.

Lets says, k=6 and m=44 and t=1

I agree till the step where you get

(2(3x)+3m)/144 = t

Now if we take 3 out as a common factor, we get

3 * ( (2x+m) / 144) = t

Since m=6, x=2. Let's plug in the values,

3 * (4+44/144) = t

3 * (1/3) = t

Notice that for t to be an integer, ((2x+m) / 144) doesn't need to be an integer. As we can see in the above example that ( (2x+m) / 144) = 1/3, and that 12 & t don't have a common factor greater than 1.

Hope that helps..

Thanks Hardik..
But a small correction from your note:

Notice that for t to be an integer, ((2x+m) / 144) doesn't need to be an integer. As we can see in the above example that ( (2x+m) / 144) = 1/3, and that 12 & t don't have a common factor greater than 1.

for t be an integer ((2x+m) / 144) [equal to t]should be an integer. But ((2x+m) / 12) [which is equal to 12t] need not be an integer.
It is late night back in India.. so I can imagine..

Thanks for the fix anyways... Have a good sleep.. Good night!!

kvcpk wrote:Thanks Hardik..
But a small correction from your note:

Notice that for t to be an integer, ((2x+m) / 144) doesn't need to be an integer. As we can see in the above example that ( (2x+m) / 144) = 1/3, and that 12 & t don't have a common factor greater than 1.

for t be an integer ((2x+m) / 144) [equal to t]should be an integer. But ((2x+m) / 12) [which is equal to 12t] need not be an integer.
It is late night back in India.. so I can imagine..

Thanks for the fix anyways... Have a good sleep.. Good night!!

I didn't understand what you are trying to say..

We have k=6 (So x=2), m=44 and t=1

In your explanation you assumed that ( (2x+m) / 144) has to be an integer. But as you can see in my previous post, ( (2x+m) / 144) = 1/3, which is not an integer, and still t is an integer.

Am I missing something?

hardik.jadeja wrote:
I didn't understand what you are trying to say..
Am I missing something?

Oops.. My mistake I think.. I mistook 2x+y to 2k+3m.. Its sleeping time here also.. Its better I sleep i think before I type some bla blah here..

Thanks for your prompt response..Good Night..

kvcpk wrote: Oops.. My mistake I think.. I mistook 2x+y to 2k+3m.. Its sleeping time here also.. Its better I sleep i think before I type some bla blah here..

Thanks for your prompt response..Good Night..

Not an issue buddy.. Happens to all of us..

That's very nice and useful conversation in the end of the day!