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

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

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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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Originally posted by Stiv on 23 Feb 2012, 02:28.
Last edited by Bunuel on 05 Feb 2019, 03:31, edited 1 time in total.
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Re: If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12  [#permalink]

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New post 23 Feb 2012, 02:40
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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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If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12  [#permalink]

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New post 26 Jun 2013, 01:39
The answer should be (A).

By simplifying the equation we get 2k + 3m = t
Does this mean that 2 and/or 3 are factors of t? Not necessarily!
Consider k=1 and m=1 => t=5. Does t have 2 and 3 as factors? No!
Alternately, consider k=3 and m=2 => t=12. In this case, t does have both k and m as factors.

Point to note:
If a positive integer is the sum of the multiples of other positive integers, it need not be a multiple of either of the integers!

Carrying on with this question,

Using statement 1: If k is a multiple of 3, then the equation can be written as
2k + 3m = t
=> 2*3n + 3m = t (where n is a positive integer)
=> 3 (2n +m) = t
=> 3 is a factor of t
=> t and 12 have a common factor greater than 1 (i.e. 3)
SUFFICIENT.

Consider statement 2: If m is a multiple of 3, we can write the equation as
2k + 3m = t
=> 2k + 3*3n = t (where n is a positive integer)
=> 2k + 9n = t
If we take n=1 and k=3, we get t=15, which has 3 as a common factor greater than 1 with 12
If we take k=1 and n=1, we get t=11, which has no common factor greater than 1 with 12
Therefore statement 2 alone is insufficient.

The answer is (A).
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Re: If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12  [#permalink]

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New post 26 Jun 2013, 02:09
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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, do t and 12  [#permalink]

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

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Re: If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12   [#permalink] 05 Feb 2019, 03:35
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