Data Sufficiency

HI Architj,

This DS question is perfect for TESTing VALUES.

We're told that X is an integer. We're asked if X is divisible by 36. This is a YES/NO question.

Fact 1: X is divisible by 12.

IF...
X = 12
The answer to the question is NO

IF....
X = 36
The answer to the question is YES
Fact 1 is INSUFFICIENT

Fact 2: X is divisible by 9

IF...
X = 9
The answer to the question is NO

IF....
X = 36
The answer to the question is YES
Fact 2 is INSUFFICIENT

Combined, we know....
X is divisible by 12
X is divisible by 9

The values of X that are divisible by BOTH numbers are multiples of 36: 36, 72, 108, etc. so the answer to the question is ALWAYS YES
Combined, SUFFICIENT

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

Architj wrote:Is the integer x divisible by 36?

(1) x is divisible by 12.
(2) x is divisible by 9.

i think the answer here should be D, but the correct is C, how?

Solution:

We are asked whether x is divisible by 36. This could also be re-written as:

Is x/36 = integer?

Statement One Alone:

x is divisible by 12.

Thus, x could be values such as 12 or 36.

We can see that when x is 12, x/36 is not an integer. That is when x is 12, x is not divisible by 36.

However, when x is 36, x/36 is an integer. That is when x is 36, x is divisible by 36.

Since we have a yes and a no answer, statement one alone is not sufficient.

Statement Two Alone:

x is divisible by 9.

Thus, x could be values such as 9 or 36.

We can see that when x is 9, x/36 is not an integer. That is when x is 9, x is not divisible by 36.

However, when x is 36, x/36 is an integer. That is when x is 36, x is divisible by 36.

Since we have a yes and a no answer, statement two alone is not sufficient.

Statements One and Two Together:

Using both statements, we know x can take values that are multiples of 9 and 12. Thus, we need to determine the least common multiple of 9 and 12.

Although there are technical ways for determining the least common multiple, the easiest method here is to analyze the multiples of 9 and 12 until we find one in common.

Starting with 9, we have: 9, 18, 27, 36, and 45

For 12, we have: 12, 24, 36

For the multiples of 12, we stopped at 36, because we see that 36 is also a multiple of 9. Thus, 36 is the least common multiple of 9 and 12, and therefore we know that 36 is the lowest possible value for x. Since x is either 36 or a multiple of 36, we know that x must be divisible by 36.

The answer is C

Architj wrote:Is the integer x divisible by 36?

(1) x is divisible by 12.
(2) x is divisible by 9.

For questions involving divisibility, 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)

We can use this strategy to REPHRASE the target question.

Original target question: Is the integer x divisible by 36?
Since 36 = (2)(2)(3)(3), we can REPHRASE the target question as follows...
REPHRASED target question: Are there two 2's and two 3's hiding in the prime factorization of x?

Statement 1: x is divisible by 12
12 = (2)(2)(3)
So, we can be certain that two 2's and one 3 are hiding in the prime factorization of x.
Of course, there COULD be more 2's and 3's hiding in the prime factorization.
So, it COULD be the case that two 2's and two 3's ARE hiding in the prime factorization of x.
Or it COULD be the case that two 2's and two 3's are NOT hiding in the prime factorization of x.
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x is divisible by 9
9 = (3)(3)
So, we can be certain that two 3's are hiding in the prime factorization of x.
Of course, there COULD be additional prime numbers hiding in the prime factorization.
So, it COULD be the case that two 2's and two 3's ARE hiding in the prime factorization of x.
Or it COULD be the case that two 2's and two 3's are NOT hiding in the prime factorization of x.
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that two 2's and one 3 are hiding in the prime factorization of x
Statement 2 tells us that two 3's are hiding in the prime factorization of x
When we combine the statements, we can be certain that two 2's and two 3's are hiding in the prime factorization of x.
Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent