If K is a common multiple of 75, 98, and 140,

[rsarashi]

If K is a common multiple of 75, 98, and 140, which of the following statements must be true?

|) k is divisible by 9.
||) k is divisible by 49.
|||) k is greater than 14,000.

A) || only
B) |||) only
C) | and || only
D) II and ||| only
E) |,||, and |||

OAD

[GMATGuruNY]

If K is a common multiple of 75, 98, and 140, which of the following statements must be true?

|) k is divisible by 9.
||) k is divisible by 49.
|||) k is greater than 14,000.

A) || only
B) |||) only
C) | and || only
D) II and ||| only
E) |,||, and |||

The prompt should indicate that k is positive.

75 = 3*5*5.
98 = 2*7*7.
140 = 2*2*5*7.

To be divisible by 75, 98 and 140, the prime-factorization of k must include the following:
2*2 (to accommodate the two 2's within 140)
3 (to accommodate the one 3 within 75)
5*5 (to accommodate the two 5's within 75)
7*7 (to accommodate the two 7's within 98).

Thus, the smallest possible value for k is as follows:
2*2*3*5*5*7*7 = (2*5)(2*5)(3)(7*7) ≈ 300*50 = 15000.
The prime-factorization in red is divisible by 75 (because it includes 3*5*5), by 98 (because it includes 2*7*7), and by 140 (because it includes 2*2*5*7).

I: k is divisible by 9
Since the prime-factorization in red includes only one 3, k does not have to be divisible by 9.
Thus, Statement I does not have to be true.
Eliminate C and E, which include Statement I.

II: k is divisible by 49
Since the prime-factorization in red includes two 7's, k must be divisible by 49.
Thus, Statement II must be true.
Eliminate B, which does not include Statement II.

III: k is greater than 14,000
The value in blue indicates that Statement III must be true.
Eliminate A, which does not include Statement III.

The correct answer is D.

[Brent@GMATPrepNow]

If K is a common multiple of 75, 98, and 140, which of the following statements must be true?

|) k is divisible by 9.
||) k is divisible by 49.
|||) k is greater than 14,000.

A) || only
B) |||) only
C) | and || only
D) II and ||| only
E) |,||, and |||

OAD

---ASIDE--------------------------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(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)
-------NOW ONTO THE QUESTION------------------------------

K is a common multiple of 75, 98, and 140
75 = (3)(5)(5). So, there is at least one 3 and two 5's hiding in the prime factorization of K
98 = (2)(7)(7). So, there is at least one 2 and two 7's hiding in the prime factorization of K
140 = (2)(2)(5)(7). So, there are at least two 2's, one 5 and one 7 hiding in the prime factorization of K
So, K = (2)(2)(3)(5)(5)(7)(7)(?)(?)(?)(?), where the ?'s represent other possible primes.

Now let's examine the statements....
|) k is divisible by 9.
9 = (3)(3)
Since we can't be sure that there are two 3's hiding in the prime factorization of K, we cannot be certain that K is divisible by 9
So, statement I is may NOT be true
ELIMINATE answer choices C and D

||) k is divisible by 49.
49 = (7)(7)
Since there are (at least) two 7's hiding in the prime factorization of K, we can be certain that K is divisible by 49
So, statement II is TRUE
ELIMINATE answer choice B

|||) k is greater than 14,000.
We already noted that K = (2)(2)(3)(5)(5)(7)(7)(?)(?)(?)(?)
Since (2)(2)(3)(5)(5)(7)(7) = 14,700, we know that K is AT LEAST as big as 14,700 .
So, we can be certain that K is greater than 14,000
So, statement III is TRUE

Answer: D

Cheers,
Brent

[Scott@TargetTestPrep]

If K is a common multiple of 75, 98, and 140, which of the following statements must be true?

|) k is divisible by 9.
||) k is divisible by 49.
|||) k is greater than 14,000.

A) || only
B) |||) only
C) | and || only
D) II and ||| only
E) |,||, and |||

OAD

Let’s prime factorize the numbers:

75 = 3 x 25 = 3 x 5^2

98 = 2 x 49 = 2 x 7^2

140 = 2 x 5 x 2 x 7 = 2^2 x 5 x 7

The least common multiple of 75, 98, and 140 is 2^2 x 3 x 5^2 x 7^2 = (2^2 x 5^2) x (3 x 7^2) = 100 x 147 = 14,700. Since k is a common multiple of 75, 98, and 140, it must be a multiple of 14,700. Now let’s look at the statements.

Statement I is not necessarily true since if k = 14,700, then it’s not divisible by 9 (notice that k has only one factor of 3).

Statement II is true because 14,700 has 7^2 = 49 as a factor, so since k is a multiple of 14,700, we see that k also has 49 as a factor, i.e., k is divisible by 49.

Statement III is true since the smallest value of k is 14,700, which is already greater than 14,000.

Answer: D

[foo]

If K is a common multiple of 75, 98, and 140, which of the following statements must be true?

|) k is divisible by 9.
||) k is divisible by 49.
|||) k is greater than 14,000.

A) || only
B) |||) only
C) | and || only
D) II and ||| only
E) |,||, and |||

OAD

Let’s prime factorize the numbers:

75 = 3 x 25 = 3 x 5^2

98 = 2 x 49 = 2 x 7^2

140 = 2 x 5 x 2 x 7 = 2^2 x 5 x 7

The least common multiple of 75, 98, and 140 is 2^2 x 3 x 5^2 x 7^2 = (2^2 x 5^2) x (3 x 7^2) = 100 x 147 = 14,700. Since k is a common multiple of 75, 98, and 140, it must be a multiple of 14,700. Now let’s look at the statements.

Statement I is not necessarily true since if k = 14,700, then it’s not divisible by 9 (notice that k has only one factor of 3).

Statement II is true because 14,700 has 7^2 = 49 as a factor, so since k is a multiple of 14,700, we see that k also has 49 as a factor, i.e., k is divisible by 49.

Statement III is true since the smallest value of k is 14,700, which is already greater than 14,000.

Answer: D

Hey @Scott@TargetTestPrep , is 0 not a multiple of every number? LCM is not 0 because by definition LCM is defined as the positive number, I guess.

[foo]

If K is a common multiple of 75, 98, and 140, which of the following statements must be true?

|) k is divisible by 9.
||) k is divisible by 49.
|||) k is greater than 14,000.

A) || only
B) |||) only
C) | and || only
D) II and ||| only
E) |,||, and |||

OAD

Let’s prime factorize the numbers:

75 = 3 x 25 = 3 x 5^2

98 = 2 x 49 = 2 x 7^2

140 = 2 x 5 x 2 x 7 = 2^2 x 5 x 7

The least common multiple of 75, 98, and 140 is 2^2 x 3 x 5^2 x 7^2 = (2^2 x 5^2) x (3 x 7^2) = 100 x 147 = 14,700. Since k is a common multiple of 75, 98, and 140, it must be a multiple of 14,700. Now let’s look at the statements.

Statement I is not necessarily true since if k = 14,700, then it’s not divisible by 9 (notice that k has only one factor of 3).

Statement II is true because 14,700 has 7^2 = 49 as a factor, so since k is a multiple of 14,700, we see that k also has 49 as a factor, i.e., k is divisible by 49.

Statement III is true since the smallest value of k is 14,700, which is already greater than 14,000.

Answer: D

Hey @Scott@TargetTestPrep , is 0 not a multiple of every number? LCM is not 0 because by definition LCM is defined as the positive number, I guess.