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Re: Which of the following could be true of at least some of the terms of [#permalink]

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12 Mar 2015, 06:50

2n denotes an even integer when n is an integer. So 2n-1 and 2n+3 are two odd integers. Looking at the answer choices II. is out of place since it's a multiple of two.

Answer is D.
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learn the rules of the game, then play better than anyone else.

2n denotes an even integer when n is an integer. So 2n-1 and 2n+3 are two odd integers. Looking at the answer choices II. is out of place since it's a multiple of two.

it is very important to answer the question very deliberately pertaining to a variable.... please remember ,it can take a value of a fraction, integer , 0 ,-ive or +ive unless specified otherwise... here we are trying to find 'atleast' so we look for any value which can satisfy the given properties.. remember 2n is not even when n =19/2.. (2n-1)=18... so ll is satisfied.. l and lll is satisfied by integer values and ll by fraction..

any value is possible with (2n-1)(2n+3).. ans E
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Re: Which of the following could be true of at least some of the terms of [#permalink]

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13 Mar 2015, 00:00

chetan2u wrote:

gmat6nplus1 wrote:

2n denotes an even integer when n is an integer. So 2n-1 and 2n+3 are two odd integers. Looking at the answer choices II. is out of place since it's a multiple of two.

it is very important to answer the question very deliberately pertaining to a variable.... please remember ,it can take a value of a fraction, integer , 0 ,-ive or +ive unless specified otherwise... here we are trying to find 'atleast' so we look for any value which can satisfy the given properties.. remember 2n is not even when n =19/2.. (2n-1)=18... so ll is satisfied.. l and lll is satisfied by integer values and ll by fraction..

any value is possible with (2n-1)(2n+3).. ans E

Hey chetan2u, you're right. I misread that n was an integer, my brain did all the rest
_________________

learn the rules of the game, then play better than anyone else.

First of all, if n = 8, then \(b_8 = (16 - 1)(16 + 3) = 15*19\)

We don’t have to calculate that: clearly, whatever it is, it is divisible by 15. Similarly, if n = 12, then \(b_{12} = (24 - 1)(24 + 3) = 23*27\)

Whatever that equals, it must be divisible by 27. Thus, I & III are true. Notice that, for any integer, 2n must be even, so both (2n – 1) and (2n + 3) are odd numbers, and their product must be odd. Every term in this sequence is an odd number. Now, no odd number can be divisible by an even number, because there is no factor of 2 in the odd number. Therefore, no terms could possibly be divisible by 18. Statement II is absolutely not true.

First of all, if n = 8, then \(b_8 = (16 - 1)(16 + 3) = 15*19\)

We don’t have to calculate that: clearly, whatever it is, it is divisible by 15. Similarly, if n = 12, then \(b_{12} = (24 - 1)(24 + 3) = 23*27\)

Whatever that equals, it must be divisible by 27. Thus, I & III are true. Notice that, for any integer, 2n must be even, so both (2n – 1) and (2n + 3) are odd numbers, and their product must be odd. Every term in this sequence is an odd number. Now, no odd number can be divisible by an even number, because there is no factor of 2 in the odd number. Therefore, no terms could possibly be divisible by 18. Statement II is absolutely not true.

Answer = (D).

Hi Magoosh Team , The question stem does not say that N is an integer . (snippet from official explaination : ' Notice that, for any integer, 2n must be even' ) i think the question is not complete. for example at \(N= \frac{1}{2}\) 15,27,18 or any other number will be able to divide \(b_n = (2n-1)*(2n+3)\)

thanks lucky
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Last edited by Lucky2783 on 21 Mar 2015, 08:10, edited 1 time in total.

First of all, if n = 8, then \(b_8 = (16 - 1)(16 + 3) = 15*19\)

We don’t have to calculate that: clearly, whatever it is, it is divisible by 15. Similarly, if n = 12, then \(b_{12} = (24 - 1)(24 + 3) = 23*27\)

Whatever that equals, it must be divisible by 27. Thus, I & III are true. Notice that, for any integer, 2n must be even, so both (2n – 1) and (2n + 3) are odd numbers, and their product must be odd. Every term in this sequence is an odd number. Now, no odd number can be divisible by an even number, because there is no factor of 2 in the odd number. Therefore, no terms could possibly be divisible by 18. Statement II is absolutely not true.

Answer = (D).

Hi Magoosh Team , The question stem does not say that N is an integer . (snippet from official explaination : ' Notice that, for any integer, 2n must be even' ) i think the question is not complete. for example at N= 1/2 15,27,18 or any other number will be able to divide \(b_n = (2n-1)*(2n+3)\)

thanks lucky

\(b_n\) is n-th term in the sequence, so n must be an integer.
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First of all, if n = 8, then \(b_8 = (16 - 1)(16 + 3) = 15*19\)

We don’t have to calculate that: clearly, whatever it is, it is divisible by 15. Similarly, if n = 12, then \(b_{12} = (24 - 1)(24 + 3) = 23*27\)

Whatever that equals, it must be divisible by 27. Thus, I & III are true. Notice that, for any integer, 2n must be even, so both (2n – 1) and (2n + 3) are odd numbers, and their product must be odd. Every term in this sequence is an odd number. Now, no odd number can be divisible by an even number, because there is no factor of 2 in the odd number. Therefore, no terms could possibly be divisible by 18. Statement II is absolutely not true.

Answer = (D).

Hi Magoosh Team , The question stem does not say that N is an integer . (snippet from official explaination : ' Notice that, for any integer, 2n must be even' ) i think the question is not complete. for example at N= 1/2 15,27,18 or any other number will be able to divide \(b_n = (2n-1)*(2n+3)\)

thanks lucky

hi lucky, although i too thought the same but i think since bn is a term in a sequence so n has to be an integer
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Sequence questions are always based on a pattern of some kind, but there might be more to the pattern than you might initially realize. It often helps to "map out" the first few terms in a sequence question, so you can "see" the numbers and spot any hidden patterns involved.

Here, we're given the 'equation' for the sequence: bN = (2N-1)(2N+3) where N is the Nth term in the sequence....

1st term = (2-1)(2+3) = (1)(5) = 5 2nd term = (4-1)(4+3) = (3)(7) = 21 3rd term = (6-1)(6+3) = (5)(9) = 45 4th term = (8-1)(8+3) = (7)(11) = 77 Etc.

From this, we can deduce a number of details: 1) EVERY term will be an ODD INTEGER 2) The sequence of "products" will work through EVERY positive odd integer (notice the 1, the 3, the 5, the 7, etc).

This helps us to prove that Roman Numerals I and III COULD be true and that II is false.

Re: Which of the following could be true of at least some of the terms of [#permalink]

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11 Mar 2016, 19:16

Bunuel wrote:

Which of the following could be true of at least some of the terms of the sequence defined by \(b_n = (2n - 1)(2n + 3)\)

I. divisible by 15 II. divisible by 18 III. divisible by 27

A. I only B. II only C. I and II only D. D. I and III only E. I, II, III

Kudos for a correct solution.

since it's a COULD be and not a MUST be true question...we can find ways for I and III to work. look at the sequence..we have 2 odd numbers multiplied. since neither of them has a factor of 2, we can eliminate choices with II right away - B, C, and E - out.

now.. for (2n-1)(2n+3), we can find values so that it would be divisible by 15 and by 27 suppose 2n+3=15. 2n=12. n=6 (12-1)(12+3)=11*15 - so yes, possible

suppose 2n+3=27 2n=24 n=12 (24-1)(24+3)=23x27 - so yes, possible.

Which of the following could be true of at least some of the terms of [#permalink]

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20 Dec 2017, 23:07

Bunuel wrote:

Lucky2783 wrote:

Hi Magoosh Team , The question stem does not say that N is an integer . (snippet from official explaination : ' Notice that, for any integer, 2n must be even' ) i think the question is not complete. for example at N= 1/2 15,27,18 or any other number will be able to divide \(b_n = (2n-1)*(2n+3)\)

thanks lucky

\(b_n\) is n-th term in the sequence, so n must be an integer.

Could n be a negative integer? May not be, I think.
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Hi Magoosh Team , The question stem does not say that N is an integer . (snippet from official explaination : ' Notice that, for any integer, 2n must be even' ) i think the question is not complete. for example at N= 1/2 15,27,18 or any other number will be able to divide \(b_n = (2n-1)*(2n+3)\)

thanks lucky

\(b_n\) is n-th term in the sequence, so n must be an integer.

Could n be a negative integer? May not be, I think.

n is an index number which indicates the position of the term in the sequence, it cannot be negative.
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