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If k, m, and t are positive integers and k/6 + m/4 = t/12 [#permalink]

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23 Feb 2012, 01:28

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54% (02:26) correct
46% (01:28) wrong based on 340 sessions

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If k, m, and t are positive integers and k/6 + m/4 = t/12, 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.

In the explanation of this question they say that the sum of two multiples of 3 give the number that is also a multiple of 3. Is that a general rule for any number? If someone can elaborate I would be grateful!

If k, m, and t are positive integers and \(\frac {k}{6} + \frac {m}{4} = \frac {t}{12}\) , 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.

In the explanation of this question they say that the sum of two multiples of 3 give the number that is also a multiple of 3. Is that a general rule for any number? If someone can elaborate I would be grateful!

If k, m, and t are positive integers and \(\frac{k}{6} + \frac{m}{4} = \frac{t}{12}\), do t and 12 have a common factor greater than 1 ?

(1) k is a multiple of 3 --> \(k=3x\), where \(x\) is a positive integer --> \(2k+3m=6x+3m=3(2x+m)=t\) --> \(t\) is multiple of 3, hence \(t\) and 12 have a common factor of 3>1. Sufficient.

(2) m is a multiple of 3 --> \(m=3y\), where \(y\) is a positive integer --> \(2k+3m=2k+9y=t\) --> \(t\) and 12 may or may not have a common factor greater than 1. Not sufficient.

Answer: A.

As for your question: If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)): Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

Re: If k, m, and t are positive integers and k/6 + m/4 = t/12 [#permalink]

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26 Jun 2013, 01:09

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Stiv wrote:

If k, m, and t are positive integers and k/6 + m/4 = t/12, 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.

In the explanation of this question they say that the sum of two multiples of 3 give the number that is also a multiple of 3. Is that a general rule for any number? If someone can elaborate I would be grateful!

We can solve the given expression and get the following

(2k+3m)/12= t/12 ------> this implies t= 2k +3 m

From St 1 we have k is a multiple of 3 so the above equation is of the form t= 2*3*a+ 3m i.e t= 6a +3m where a is a positive integer (since K is a positive integer "a" cannot be zero)

thus t = 3( 2a+m) if a =1, m=1 then t= 9 ; an 9 and 12 have 3 as common factor other than 1 similarly if a=2, m=1 we have t=15, and both 15 and 12 have 3 as common factor since t has 3 as one of its factors and 12 also has 3 as one of its factor and therefore "t" and 12 will always have 3 as a factor other than 1

from St2 we have t= 2k+ 3*3b -----> t= 2k+9b where b is a positive integer

Here if k=1 and b =1, then t= 11; 11 and 12 do not have any common factor other than 1 but if k=3 and b=3 then we have t= 24 ; 24 and 12 have many common factor

therefore ans should be A
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Re: If k, m, and t are positive integers and k/6 + m/4 = t/12 [#permalink]

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01 Feb 2015, 04:33

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Re: If k, m, and t are positive integers and k/6 + m/4 = t/12 [#permalink]

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16 Mar 2016, 05:53

Superb QUESTION Here we need to write k as 3*p for some integer p so 3 must be in the GCD hence A is sufficient AS for statement 2 => t=5=> NO for t=10=> YES hence not sufficient hence A
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