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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] New post   Updated on: 05 Feb 2019, 03:31
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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.


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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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A

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

    Simplify \(\frac{k}{6} + \frac{m}{4} = \frac{t}{12}\) by multiplying by 12:

    \(t=2k+3m\).

(1) k is a multiple of 3:

    \(k=3x\), where \(x\) is a positive integer;

    \(t=2k+3m=6x+3m=3(2x+m)\);

    \(t\) is multiple of 3, hence \(t\) and 12 have a common factor of 3, which is greater than 1.


Sufficient.

(2) m is a multiple of 3:

    \(m=3y\), where \(y\) is a positive integer;

    \(t=2k+3m=2k+9y\);

    \(t\) and 12 may or may not have a common factor greater than 1.


Not sufficient.

Answer: A.

As for your question:

Stiv wrote:
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 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] New post   26 Jun 2013, 01:39
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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] 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] New post   16 Mar 2016, 06:53
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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Re: If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12 [#permalink] New post   02 Mar 2020, 14:19
Quote:
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.

Hello,
IanStewart
So, it seems that we need the value of \(t\) is equals to any prime number to have the answer NO in statement 2, right?
Thanks__
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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] New post   02 Mar 2020, 18:27
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Asad wrote:
Hello,
IanStewart
So, it seems that we need the value of \(t\) is equals to any prime number to have the answer NO in statement 2, right?
Thanks__


To get a 'no' answer using Statement 2, you just need t to equal something that doesn't share any factors (besides 1) with 12. So t could be 25 or 35, say; it doesn't need to be prime, though making t equal to a small prime like 5 or 7 is a very good choice if you're testing numbers. And t also can't be just any prime -- if t were 2 or 3, then you would not get a 'no' answer to the question, though it's impossible for t to equal 2 or 3 anyway in this equation, if m and k are positive integers.
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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] New post   06 Mar 2021, 22:01
Bunuel wrote:
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.


Totally missed this here. How do we get for statement 1:
6x+3m=3(2x+m)=t

e.g. Where did the x variable come from?

Also wondering why the original equation simplified to this:
2k+3m = t

How come it's not 3k + 2m = t? (If we were to multiply each term by 12)
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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] New post   07 Mar 2021, 02:37
Expert Reply
CEdward wrote:
Bunuel wrote:
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.


Totally missed this here. How do we get for statement 1:
6x+3m=3(2x+m)=t

e.g. Where did the x variable come from?

Also wondering why the original equation simplified to this:
2k+3m = t

How come it's not 3k + 2m = t? (If we were to multiply each term by 12)


1. k is a multiple of 3 means that \(k=3x\) for some positive integer \(x\). For example, when x = 1, then k = 3, when x = 2, then k = 6 and so on. Next, if you substitute k with 3x in 2k + 3m you'll get 6x + 3m.

2. \(\frac{k}{6} + \frac{m}{4} = \frac{t}{12}\) --> multiply both sides by 12 to get \(2k+3m=t\).
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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] New post   09 Dec 2023, 17:10
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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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