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

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:



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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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).

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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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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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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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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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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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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