BlueDragon2010 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
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
