DS - Number Properties

[f2001290]

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.

[sanju09]

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]

[mj78ind]

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

[kvcpk]

I believe Answer should be A.

What is the source of this question?

[sanju09]

I believe Answer should be A.

What is the source of this question?

please go ahead

[kvcpk]

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.

[hardik.jadeja]

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..

[kvcpk]

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!!

[hardik.jadeja]

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?

[kvcpk]

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..

[hardik.jadeja]

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..

[sanju09]

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