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

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

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New post 23 Feb 2012, 01:28
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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!

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Re: General Math Question  [#permalink]

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New post 23 Feb 2012, 01:40
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Stiv wrote:
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 ?

\(\frac{k}{6} + \frac{m}{4} = \frac{t}{12}\) --> \(2k+3m=t\).

(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.

Hope it's clear.
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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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New post 26 Jun 2013, 00:39
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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New post 25 Nov 2018, 23:14
Could someone kindly correct or reinforce my logic below? I struggled to understand how statement (b) was incorrect, but as I typed this post out I think i've got it:

T and 12 can have either 2, 4 or 3 as a common prime factor.

If T = 2k + 3m and m=multiple of 3 then lets test values for k and m

k=0 and m=3
T= 2(0) + 3(3)
T = 9 therefore T and 12 share GCF of 3

k=1 and m=3
T= 2(1) + 3(3)
T= 2 + 9
T = 11

T and 12 share no common factors. (consecutive integer rule) GCF = 1

Since there are 2 possible answers --(b)-->IS
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If k, m, and t are positive integers and k/6 + m/4 = t/12 &nbs [#permalink] 25 Nov 2018, 23:14
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