A in Answer.
t=(2k+3m)
Let k=3A
t=3*(2A+m)
Let m=3B
t=(2k+9B)
prime & divisibility
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rajeshsinghgmat
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Excellent explanation. Big Fan !!GMATGuruNY wrote:To clear the fractions, multiple the given equation by 12:Gurpinder 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.
12(k/6 + m/4) = 12/(t/12)
2k + 3m = t.
The factors of 12 greater than 1 are 2,3,4,6, and 12.
Question rephrased: Is t a multiple of 2,3,4,6 or 12?
Statement 1: k is a multiple of 3
Thus:
t = 2(multiple of 3) + 3m.
t = multiple of 3 + multiple of 3.
Since the sum of two multiples of 3 must also be a multiple of 3, t is a multiple of 3.
SUFFICIENT.
Statement 2: m is a multiple of 3
If k=3 and m=3, then t = 2(3) + 3(3) = 15.
In this case, t is a multiple of 3.
If k=1 and m=3, then t = 2(1) + 3(3) = 11.
In this case, t is not a multiple of 2,3,4,6, or 12.
INSUFFICIENT.
The correct answer is A.
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Brent@GMATPrepNow
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Target question: Do t and 12 have a common factor (divisor) greater than 1?
A little background information
For questions involving factors (aka "divisors"), we can say:
If k is a divisor of N, then k is "hiding" within the prime factorization of N
Examples:
3 is a divisor of 24 <--> 24 = (2)(2)(2)(3)
5 is a divisor of 70 <--> 70 = (2)(5)(7)
8 is a divisor of 56 <--> 56 = (2)(2)(2)(7)
Similarly, for questions involving multiples, we can say:
If N is a multiple of k, then k is "hiding" within the prime factorization of N
Examples:
24 is a multiple of 3 <--> 24 = (2)(2)(2)(3)
70 is a multiple of 5 <--> (2)(5)(7)
330 is a multiple of 6 <--> 330 = (2)(3)(5)(11)
Okay, now let's solve the question....
Given: (k/6)+ (m/4) = (t/12)
Simplify this by multiplying both sides by 12 to get 2k + 3m = t
Statement 1: k is a multiple of 3.
In other words, 3 is hiding in the prime factorization of k
So, we know that k =(3)(?)(?)(?)...
Aside: notice that the prime factorization of k may or may not have any primes other than the 3. All we can be certain of is that there is one 3 within the prime factorization (thus the question marks in the factorization)
From here, we'll take our given information, 2k + 3m = t, and replace k with (3)(?)(?)(?) to get: (2)(3)(?)(?)(?) + 3m = t
At this point, we can factor out a 3 to get: 3[(2)(?)(?)(?) + m] = t, which means 3 is a divisor of t.
Since 3 is also a divisor of 12, we can see that t and 12 have a common factor (divisor) greater than 1.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: m is a multiple of 3.
In other words, 3 is hiding in the prime factorization of m
So, we know that m =(3)(?)(?)(?)...
From here, we'll take our given information, 2k + 3m = t, and replace m with (3)(?)(?)(?) to get: 2k + 3(3)(?)(?)(?) = t
At this point, we cannot factor out any number from the expression.
So, we cannot determine whether t has any divisors (factors) in common with 12
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Aside:
To demonstrate that statement 2 is NOT SUFFICIENT, consider these two contradictory sets of values for k, m and t.
case a: k=3, m=3, t=15, in which case t and 12 have a common factor (divisor) greater than 1.
case b: k=1, m=3, t=11, in which case t and 12 do not have a common factor (divisor) greater than 1.
Cheers,
Brent
By the way, here's another question we can solve using prime factorization: https://www.beatthegmat.com/divisibility-t91797.html
A little background information
For questions involving factors (aka "divisors"), we can say:
If k is a divisor of N, then k is "hiding" within the prime factorization of N
Examples:
3 is a divisor of 24 <--> 24 = (2)(2)(2)(3)
5 is a divisor of 70 <--> 70 = (2)(5)(7)
8 is a divisor of 56 <--> 56 = (2)(2)(2)(7)
Similarly, for questions involving multiples, we can say:
If N is a multiple of k, then k is "hiding" within the prime factorization of N
Examples:
24 is a multiple of 3 <--> 24 = (2)(2)(2)(3)
70 is a multiple of 5 <--> (2)(5)(7)
330 is a multiple of 6 <--> 330 = (2)(3)(5)(11)
Okay, now let's solve the question....
Given: (k/6)+ (m/4) = (t/12)
Simplify this by multiplying both sides by 12 to get 2k + 3m = t
Statement 1: k is a multiple of 3.
In other words, 3 is hiding in the prime factorization of k
So, we know that k =(3)(?)(?)(?)...
Aside: notice that the prime factorization of k may or may not have any primes other than the 3. All we can be certain of is that there is one 3 within the prime factorization (thus the question marks in the factorization)
From here, we'll take our given information, 2k + 3m = t, and replace k with (3)(?)(?)(?) to get: (2)(3)(?)(?)(?) + 3m = t
At this point, we can factor out a 3 to get: 3[(2)(?)(?)(?) + m] = t, which means 3 is a divisor of t.
Since 3 is also a divisor of 12, we can see that t and 12 have a common factor (divisor) greater than 1.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: m is a multiple of 3.
In other words, 3 is hiding in the prime factorization of m
So, we know that m =(3)(?)(?)(?)...
From here, we'll take our given information, 2k + 3m = t, and replace m with (3)(?)(?)(?) to get: 2k + 3(3)(?)(?)(?) = t
At this point, we cannot factor out any number from the expression.
So, we cannot determine whether t has any divisors (factors) in common with 12
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Aside:
To demonstrate that statement 2 is NOT SUFFICIENT, consider these two contradictory sets of values for k, m and t.
case a: k=3, m=3, t=15, in which case t and 12 have a common factor (divisor) greater than 1.
case b: k=1, m=3, t=11, in which case t and 12 do not have a common factor (divisor) greater than 1.
Cheers,
Brent
By the way, here's another question we can solve using prime factorization: https://www.beatthegmat.com/divisibility-t91797.html
Brent Hanneson - Creator of GMATPrepNow.com
