If s is an integer, is 24 a divisor of s ?

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If s is an integer, is 24 a divisor of s ?

(1) Each of the numbers 3 and 8 is a divisor of s.
(2) Each of the numbers 4 and 6 is a divisor of s.




OA A

Source: Official Guide

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by [email protected] » Wed Jan 01, 2020 12:48 am
BTGmoderatorDC wrote:If s is an integer, is 24 a divisor of s ?

(1) Each of the numbers 3 and 8 is a divisor of s.
(2) Each of the numbers 4 and 6 is a divisor of s.

OA A

Source: Official Guide
Let's take each statement one by one.

(1) Each of the numbers 3 and 8 is a divisor of s.

Since 3 and 8 are co-prime, 3*8 = 24 must be a divisor of s. Sufficient.

(2) Each of the numbers 4 and 6 is a divisor of s.

Since 4 and 6 are not co-prime, their LCM is 12. Thus, though s is divisible by 12, it may/may not be divisible by 24. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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BTGmoderatorDC wrote:
Tue Dec 31, 2019 4:06 am
If s is an integer, is 24 a divisor of s ?

(1) Each of the numbers 3 and 8 is a divisor of s.
(2) Each of the numbers 4 and 6 is a divisor of s.
OA A
Source: Official Guide
------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 k is a divisor of N, then k is "hiding" within the prime factorization of N

Consider these examples:
3 is a divisor of 24, because 24 = (2)(2)(2)(3), and we can clearly see the 3 hiding in the prime factorization.
Likewise, 5 is a divisor of 70 because 70 = (2)(5)(7)
And 8 is a divisor of 112 because 112 = (2)(2)(2)(2)(7)
And 15 is a divisor of 630 because 630 = (2)(3)(3)(5)(7)
------ONTO THE QUESTION----------------------------
24 = (2)(2)(2)(3)
So, we can rephrase the target question as....
REPHRASED target question: Are there three 2's and one 3 "hiding" in the prime factorization of s?

Statement 1: Each of the numbers 3 and 8 is a divisor of s.
This tells us that 3 is hiding in the prime factorization of s
And, since 8 = (2)(2)(2), we also now know that three 2's are hiding in the prime factorization of s
So, the answer to the REPHRASED target question is YES, there three 2's and one 3 "hiding" in the prime factorization of s
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: Each of the numbers 4 and 6 is a divisor of s
Since 4 = (2)(2), we now know that two 2's are hiding in the prime factorization of s
Since 6 = (2)(3), we now know that one 2 and one 3 are hiding in the prime factorization of s
So, all we can be certain of is that there are two 2's and one 3 hiding in the prime factorization of s
Consider these two possible cases:
Case a: s = 12, in which case 24 is NOT a divisor of s
Case b: s = 24, in which case 24 IS a divisor of s
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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