Divisibility

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by Brent@GMATPrepNow » Sun Oct 13, 2013 6:01 pm
[email protected] wrote:For some integer q, q^2 - 5 is divisible by all of the following EXCEPT

(A) 29
(B) 30
(C) 31
(D) 38
(E) 41

Answer is B
There's a nice rule that goes something like this:
If A is divisible by k, but B is not divisible by k, then A+B is NOT divisible by k, and A-B is NOT divisible by k.

There's another rule that says, "If there are n consecutive integers, then 1 of them is divisible by n"

Okay, now onto the question.

IMPORTANT: Notice that only answer choice B is divisible by 3. We'll use this fact in our solution.

Rewrite the given expression: q² - 5 = q² - 1 - 4 = (q - 1)(q + 1) - 4

NOTICE that (q - 1), q and (q + 1) are 3 consecutive integers, so one of them must be divisible by 3 (from the red rule above). So, let's consider 3 possible cases:

(q - 1) is divisible by 3
So, (q - 1)(q + 1) must also be divisible by 3
Since 4 is NOT divisible by 3, we know that (q - 1)(q + 1) - 4 is NOT divisible by 3 (from the green rule above)
If (q - 1)(q + 1) - 4 is not divisible by 3, it CANNOT be divisible by 30

q is divisible by 3
So, q² must also be divisible by 3
Since 5 is NOT divisible by 3, we know that q² - 5 is NOT divisible by 3 (from the green rule above)
If q² - 5 is not divisible by 3, then it CANNOT be divisible by 30

(q + 1) is divisible by 3
So, (q - 1)(q + 1) must also be divisible by 3
Since 4 is NOT divisible by 3, we know that (q - 1)(q + 1) - 4 is NOT divisible by 3 (from the green rule above)
If (q - 1)(q + 1) - 4 is not divisible by 3, it CANNOT be divisible by 30

Now that we've examined all 3 possible cases, we can conclude that q² - 5 CANNOT be divisible by 30

Answer = B

Cheers,
Brent
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by Brent@GMATPrepNow » Sun Oct 13, 2013 6:02 pm
[email protected] wrote:For some integer q, q^2 - 5 is divisible by all of the following EXCEPT

(A) 29
(B) 30
(C) 31
(D) 38
(E) 41

Answer is B
Another approach is to plug in numbers and eliminate answer choices.

q = 6: q² - 5 = 31 ELIMINATE C
q = 7: q² - 5 = 44
q = 8: q² - 5 = 59
q = 9: q² - 5 = 76 = (2)(38) ELIMINATE D
q = 10: q² - 5 = 95
q = 11: q² - 5 = 116 = (4)(29) ELIMINATE A
q = 12: q² - 5 = 139
q = 13: q² - 5 = 164 = (4)(41) ELIMINATE E
STOP

The correct answer must be B

Cheers,
Brent
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by [email protected] » Sun Oct 13, 2013 6:08 pm
But im not sure why we use divisibility by 3 as the starting point!
Brent@GMATPrepNow wrote:
[email protected] wrote:For some integer q, q^2 - 5 is divisible by all of the following EXCEPT

(A) 29
(B) 30
(C) 31
(D) 38
(E) 41

Answer is B
There's a nice rule that goes something like this:
If A is divisible by k, but B is not divisible by k, then A+B is NOT divisible by k, and A-B is NOT divisible by k.

There's another rule that says, "If there are n consecutive integers, then 1 of them is divisible by n"

Okay, now onto the question.

IMPORTANT: Notice that only answer choice B is divisible by 3. We'll use this fact in our solution.

Rewrite the given expression: q² - 5 = q² - 1 - 4 = (q - 1)(q + 1) - 4

NOTICE that (q - 1), q and (q + 1) are 3 consecutive integers, so one of them must be divisible by 3 (from the red rule above). So, let's consider 3 possible cases:

(q - 1) is divisible by 3
So, (q - 1)(q + 1) must also be divisible by 3
Since 4 is NOT divisible by 3, we know that (q - 1)(q + 1) - 4 is NOT divisible by 3 (from the green rule above)
If (q - 1)(q + 1) - 4 is not divisible by 3, it CANNOT be divisible by 30

q is divisible by 3
So, q² must also be divisible by 3
Since 5 is NOT divisible by 3, we know that q² - 5 is NOT divisible by 3 (from the green rule above)
If q² - 5 is not divisible by 3, then it CANNOT be divisible by 30

(q + 1) is divisible by 3
So, (q - 1)(q + 1) must also be divisible by 3
Since 4 is NOT divisible by 3, we know that (q - 1)(q + 1) - 4 is NOT divisible by 3 (from the green rule above)
If (q - 1)(q + 1) - 4 is not divisible by 3, it CANNOT be divisible by 30

Now that we've examined all 3 possible cases, we can conclude that q² - 5 CANNOT be divisible by 30

Answer = B

Cheers,
Brent

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by Brent@GMATPrepNow » Sun Oct 13, 2013 6:19 pm
[email protected] wrote:But im not sure why we use divisibility by 3 as the starting point!
Good question.
First, we can rewrite q² - 5 as (q - 1)(q + 1) - 4
Next, notice that (q - 1), q, and (q + 1) are three consecutive integers (we know something about 3 consecutive integers)
Finally, I recognized that all of the answer choices are divisible by 3 EXCEPT B

For these reasons, I explored whether or not q² - 5 is divisible by 3.

Cheers,
Brent
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