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On combining both statements, we get that a is divisible by 3 and 7, b is divisible 5and7, and since they are consecutive, ab is divisible by 2 - which leads to ab being divisible by 30. But if they are consecutive integers, how can both a and b be divisible by 7? Doesn't that lead to the conclusion that there is no such a and ab, and hence that x is not an integer?

Re: If a and b are consecutive positive integers, and ab=30x [#permalink]

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09 Nov 2009, 22:30

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4test1 wrote:

If a and b are consecutive positive integers, and \(ab = 30x\), is x a non-integer?

(1) a^2 is divisible by 21 (2) 35 is a factor of b^2

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient. (B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient. (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. (D) Each statement ALONE is sufficient. (E) Statements (1) and (2) TOGETHER are NOT sufficient.

On combining both statements, we get that a is divisible by 3 and 7, b is divisible 5and7, and since they are consecutive, ab is divisible by 2 - which leads to ab being divisible by 30. But if they are consecutive integers, how can both a and b be divisible by 7? Doesn't that lead to the conclusion that there is no such a and ab, and hence that x is not an integer?

Yes/No

1. \(a^2/(21)\) = \(a^2/(7*3) = int\) not sufficient. 2. \(b^2/(35)\) = \(b^2/(7*5) = int\) not sufficient.

Re: If a and b are consecutive positive integers, and ab=30x [#permalink]

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09 Nov 2009, 22:42

This is a tricky question. I’ll give it a shot. Each statement individually is not sufficient.

(1) a^2 is divisible by 21 This statement tells us that 21 is a factor of a^2. This means that sqrt(21) is a factor of a, or sqrt(3x7) is a factor of a. Does not give us enough info.

(2) 35 is a factor of b^2 This tells us sqrt(35) or sqrt(5x7) is a factor of b. Does not give us enough info.

(1 and 2 together) Sqrt(3x7) * sqrt(5x7) is a factor of ab. So that means that sqrt(3x5) * 7 is a factor of AB. The fact that ab and b are consecutive does in fact imply one of them is even meaning the number is also divisible by 2. However, we do not know if it is sufficient as we know the only non-integer factors of ab are 7 and 2.

Hence C is correct.

Edit: I didn’t state this explicitly before but either individual statement does not give us any information to determine non-integer factors. I am factoring it by non-integers, the square roots, to show how I arrive at the final conclusion.

Re: If a and b are consecutive positive integers, and ab=30x [#permalink]

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09 Nov 2009, 23:07

The positive integers are not each individually divisible by 7 (and no this is not a possibility). The multiple of the positive integers is divisible by 7, implying only one of either a or b in ab needs to be divisible by 7.

Re: If a and b are consecutive positive integers, and ab=30x [#permalink]

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10 Nov 2009, 00:28

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Correct me if I'm wrong but here is my logic:

Neither of the statements say A or B are divisible by 7. The first one says A^2 is divisible by 21. \(\frac{A^2}{21} = N\) where N is some whole number. \(A^2 = 21N\) \(A = \sqrt{21N}\) since A is positive. \(A = \sqrt{3} * \sqrt{7} * N\) No factor of A is 7. However\(\sqrt{7}\) is a factor of A.

Re: If a and b are consecutive positive integers, and ab=30x [#permalink]

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18 Nov 2009, 16:57

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how about this way: ab=30x, is x an integer?

ab/30 = x

30 factors into 2,3,5, therefore if x has to be an integer, then the product ab has to consist of 2,3,5 in it.

1) a^2 divisible by 21 therefore a contains at the very least 7,3 not sufficient by itself since we do not know about 2 and 5 2) b^2 divisible by 35 therefore b contains at the very least 7, 5 not sufficient by itself

combine 1 and 2 ab contains 7,7,3,5 at the very least AND ab also contains a 2, since they are consecutive integers now we know that the product ab contains at least one set of 2,3,5

therefore ab divided by 30 is an integer, or x is an integer.... answer C reasoning sounds right?

Re: If a and b are consecutive positive integers, and ab=30x [#permalink]

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18 Nov 2009, 18:24

Quote:

Correct me if I'm wrong but here is my logic:

Neither of the statements say A or B are divisible by 7. The first one says A^2 is divisible by 21. \frac{A^2}{21} = N where N is some whole number. A^2 = 21N A = \sqrt{21N} since A is positive. A = \sqrt{3} * \sqrt{7} * N No factor of A is 7. However\sqrt{7} is a factor of A.

The above won't hold true. It's given that a is an integer. So A cannot be divisible by \(sqrt(7)\) - it has to be divisible by 7. Same for b.

So if a,b are consecutive integers, both cannot be divisible by 7. I'll go with C but it doesn't seem possible.

Re: If a and b are consecutive positive integers, and ab=30x [#permalink]

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18 Nov 2009, 19:18

4test1 wrote:

The above won't hold true. It's given that a is an integer. So A cannot be divisible by sqrt(7) it has to be divisible by 7.

Why can’t A be divisible by sqrt(7)? Integers can still be divided by non integers, such as irrational numbers or fractions. For example: If A = 7 = sqrt(7) * sqrt(7) = 7/3 * 3/7

Perhaps we need a maths guru such as Bunuel to clear this up.

Neither of the statements say A or B are divisible by 7. The first one says A^2 is divisible by 21. \frac{A^2}{21} = N where N is some whole number. A^2 = 21N A = \sqrt{21N} since A is positive. A = \sqrt{3} * \sqrt{7} * N No factor of A is 7. However\sqrt{7} is a factor of A.

The above won't hold true. It's given that a is an integer. So A cannot be divisible by \(sqrt(7)\) - it has to be divisible by 7. Same for b.

So if a,b are consecutive integers, both cannot be divisible by 7. I'll go with C but it doesn't seem possible.

I agree with 4test1.

This question seems to be wrong.

\(a^2=21p\) --> \(a=\sqrt{21p}\) --> as \(p\) is an integer too it should be of a type of \(p=21n^2\), for \(a\) to be an integer. --> \(a=\sqrt{21*21n^2}=21n\) --> \(a\) is a multiple of \(21\).

Same for \(b\), \(b=35m\).

We got that \(a\) is a multiple of \(21\), and \(b\) is a multiple of \(35\). So they share the same factor \(7\). But consecutive integers are co-prime they don't share any common factor but 1.

Statements contradict.

What is the source of this question?
_________________

Re: If a and b are consecutive positive integers, and ab=30x [#permalink]

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19 Nov 2013, 09:00

Bunuel wrote:

4test1 wrote:

Quote:

Correct me if I'm wrong but here is my logic:

Neither of the statements say A or B are divisible by 7. The first one says A^2 is divisible by 21. \frac{A^2}{21} = N where N is some whole number. A^2 = 21N A = \sqrt{21N} since A is positive. A = \sqrt{3} * \sqrt{7} * N No factor of A is 7. However\sqrt{7} is a factor of A.

The above won't hold true. It's given that a is an integer. So A cannot be divisible by \(sqrt(7)\) - it has to be divisible by 7. Same for b.

So if a,b are consecutive integers, both cannot be divisible by 7. I'll go with C but it doesn't seem possible.

I agree with 4test1.

This question seems to be wrong.

\(a^2=21p\) --> \(a=\sqrt{21p}\) --> as \(p\) is an integer too it should be of a type of \(p=21n^2\), for \(a\) to be an integer. --> \(a=\sqrt{21*21n^2}=21n\) --> \(a\) is a multiple of \(21\). for \(b\), \(b=35m\).

We got that \(a\) is a multiple of \(21\), and \(b\) is a multiple of \(35\). So they share the same factor \(7\). But consecutive integers are co-prime they don't share any common factor but 1.

Statements contradict.

What is the source of this question?

The OP has the problem listed incorrectly, it's from Manhattan's advanced Quant book.

The question is as follows:

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 \(b^2\).

gmatclubot

Re: If a and b are consecutive positive integers, and ab=30x
[#permalink]
19 Nov 2013, 09:00

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