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Divisibility

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Divisibility

by knight247 » Tue Jan 17, 2012 1:55 am
If a and b are consecutive positive integers, and ab=30x, is x an integer?
(1)a² is divisible by 25
(2)63 is a factor of b²

OA is C
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Source: — Data Sufficiency |
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by shankar.ashwin » Tue Jan 17, 2012 5:44 am
Is ab/2*3*5 = integer?

given a and b are consecutive integers.

(1) a² is a factor of 25, therefore 'a' is a factor of 5. But we dont know if 'a' contains a 2 or 3- Insuff

(2) b² is a factor of 7*3*3, therefore 'b' should be a factor of 7 and 3 (21) - Dont know if it contains a 5 - Insuff.

Together, since numbers are consecutive one of them should be even (factor of 2). 'a' contains a 5 and 'b' contains a '3'. Therefore ab will be divisible by 30 - C

PS: Possible values of ab could be 20*21 which is divisible by 30.
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by GMATGuruNY » Tue Jan 17, 2012 7:10 am
knight247 wrote:If a and b are consecutive positive integers, and ab=30x, is x an integer?
(1)a² is divisible by 25
(2)63 is a factor of b²

OA is C
It is given that x = ab/30.
Thus, in order for x to be an integer, ab must be a multiple of 30.
Since 30=2*3*5, in order for ab to be a multiple of 30, it must be a multiple of 2, 3, and 5.
Since a and b are consecutive integers, one of them is even and thus divisible by 2.
Thus, we know that ab is divisible by 2; what we don't know is whether ab is divisible by 3 and 5.

Question rephrased: Is ab a multiple of 3 and 5?

Statement 1: a² is divisible by 25.
25 = 5*5.
In order for a² to be multiple of 5*5, a must be a multiple of 5.
No information about whether ab is a multiple of 3.
INSUFFICIENT.

Statement 2: 63 is a factor of b².
63 = 3*3*7.
In order for b² to be multiple of 3*3*7, b must be a multiple of 3 and 7.
No information about whether ab is a multiple of 5.
INSUFFICIENT.

Statements combined:
Since a is a multiple of 5 and b is a multiple of 3, ab is a multiple of 3 and 5.
SUFFICIENT.

The correct answer is C.

For the curious: a=20, b=21 and a=125, b=126 satisfy all the conditions in the problem.
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