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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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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. 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
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23 Feb 2012, 02:40
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 \(ab=3\), 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 \(ab=1\), 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 \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), 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
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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
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26 Jun 2013, 02:09
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
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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
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05 Feb 2019, 03:35
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Re: If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12
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