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If a and b are consecutive positive integers, and ab = 30x
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30 Nov 2013, 19:37
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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
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Re: If a and b are consecutive positive integers, and ab = 30x
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02 Dec 2013, 02:00
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.
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Re: If a and b are consecutive positive integers, and ab = 30x
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01 Dec 2013, 13:45
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.
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Re: If a and b are consecutive positive integers, and ab = 30x
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19 Jun 2015, 08:21
I have one doubt. If b^2 has a factor 63 Then, b^2 has factors 3, 3, 7 for sure atleast once. Then, how can we be so sure that b has also 3 and 7 as factors. isnt b has only 3 and \sqrt{7} as factors? Bunuel wrote: 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.



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Re: If a and b are consecutive positive integers, and ab = 30x
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19 Jun 2015, 08:31
mpingo wrote: I have one doubt. If b^2 has a factor 63 Then, b^2 has factors 3, 3, 7 for sure atleast once. Then, how can we be so sure that b has also 3 and 7 as factors. isnt b has only 3 and \sqrt{7} as factors? Bunuel wrote: 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. Hi, b is a positive integer so it cannot be3*\(\sqrt{7}\)* something else... therefore it has to be a multiple of 7 and not \(\sqrt{7}\)
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Re: If a and b are consecutive positive integers, and ab = 30x
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19 Jun 2015, 08:31
mpingo wrote: I have one doubt. If b^2 has a factor 63 Then, b^2 has factors 3, 3, 7 for sure atleast once. Then, how can we be so sure that b has also 3 and 7 as factors. isnt b has only 3 and \sqrt{7} as factors? Bunuel wrote: 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. 1. Only positive integers are considered as factors. So, saying that \(\sqrt{7}\) is a factor of some integer does not make sense. 2. Exponentiation does not "produce" primes, meaning that for positive integers m and n, m^n will have the same exact prime factors as m itself does. Thus, if 3 and 7 are factors of b^2 it must be true that both of them are factors of b too (else how would they appear in b^2). Similar questions to practice: ifxisanintegerisx3divisibleby165973.htmlhowmanydifferentprimenumbersarefactorsofthepositive126744.htmlifn2nyieldsanintegergreaterthan0isndivisibleby126648.htmlifkisapositiveintegeriskthesquareofaninteger55987.htmlifkisapositiveintegerhowmanydifferentprimenumbers95585.htmlifnistheintegerwhether30isafactorofn126572.htmlifxisanintegerisx21x5anevennumber104275.htmlHope it helps.
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Re: If a and b are consecutive positive integers, and ab = 30x
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