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Maxirosario2012
GMAT 1: 680 Q49 V34
Posts: 51
30 Nov 2013, 19:37
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C
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If a and b are consecutive positive integers, and ab = 30x, is x an integer?
(1) a^2 is divisible by 25
(2) 63 is a factor of b^2
(1) a^2 is divisible by 25
(2) 63 is a factor of b^2
02 Dec 2013, 02:00
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If a and b are consecutive positive integers, and ab = 30x, is x an integer?
\(ab = 30x\) --> \(x=\frac{ab}{30}\). x to be an integer ab must be a multiple of 30.
(1) a^2 is divisible by 25. This implies that a is a multiple of 5. If a=5 and b=4, then ab=20 and x is NOT an integer but if a=5 and b=6, then ab=30 and x IS an integer. Not sufficient.
(2) 63 is a factor of b^2 --> \(63=3^2*7\). Thus b is a multiple of 3 (and 7). If b=21 and a=22, then ab=21*22 and x is NOT an integer but if b=21*30, then ab={a multiple of 30} and x IS and integer. Not sufficient.
(1)+(2) We have that a is a multiple of 5 and b is a multiple of 3. Also, since they are consecutive integers, then either one must be even. Therefore ab is a multiple of 2*3*5=30. Sufficient.
Answer: C.
Hope it's clear.
\(ab = 30x\) --> \(x=\frac{ab}{30}\). x to be an integer ab must be a multiple of 30.
(1) a^2 is divisible by 25. This implies that a is a multiple of 5. If a=5 and b=4, then ab=20 and x is NOT an integer but if a=5 and b=6, then ab=30 and x IS an integer. Not sufficient.
(2) 63 is a factor of b^2 --> \(63=3^2*7\). Thus b is a multiple of 3 (and 7). If b=21 and a=22, then ab=21*22 and x is NOT an integer but if b=21*30, then ab={a multiple of 30} and x IS and integer. Not sufficient.
(1)+(2) We have that a is a multiple of 5 and b is a multiple of 3. Also, since they are consecutive integers, then either one must be even. Therefore ab is a multiple of 2*3*5=30. Sufficient.
Answer: C.
Hope it's clear.
01 Dec 2013, 13:45
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In order for ab = 30x = 2*3*5*x, ab needs to be divisible by 2,3 and 5 (because x can be any positive number from 1, 2...)
-a and b are consecutive positive integer, so one must be even and the other is odd, so ab will be divisible by 2.
-a^2 is divisible by 25 = 5^2, so a is divisible by 5.
-63 = 3^2 * 7 is a factor of b^2, so b is divisible by 3 and 7.
As you can see, the condition a alone is not sufficient because we only know ab will be divisible by 2 and 5. Condition b alone is not sufficient also because we only know ab will be divisible by 2, 3 and 7. If a and b together, we know ab = 2*5*3*7*something, so sufficient.
Therefore, the correct answer is C.
Hope this help!
-a and b are consecutive positive integer, so one must be even and the other is odd, so ab will be divisible by 2.
-a^2 is divisible by 25 = 5^2, so a is divisible by 5.
-63 = 3^2 * 7 is a factor of b^2, so b is divisible by 3 and 7.
As you can see, the condition a alone is not sufficient because we only know ab will be divisible by 2 and 5. Condition b alone is not sufficient also because we only know ab will be divisible by 2, 3 and 7. If a and b together, we know ab = 2*5*3*7*something, so sufficient.
Therefore, the correct answer is C.
Hope this help!
