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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