Which of the following could be true of at least some of the terms of

Question

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. I and III only
E. I, II, III

gmat6nplus1 · Answer

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.

chetan2u · Answer

hi  gmat6nplus1 ,

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

gmat6nplus1 · Answer

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

23a2012 · Answer

the answer is E
where N could be any positive number, negative number or zero

Bunuel · Answer

MAGOOSH OFFICIAL SOLUTION:

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).

Lucky2783 · Answer

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

Bunuel · Answer

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

chetan2u · Answer

hi lucky,
although i too thought the same but i think since b n  is a term in a sequence so n has to be an integer

Lucky2783 · Answer

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

i see. thanks  Bunuel  .

Lucky2783 · Answer

yeah , we have to be very careful in reading the question.

EMPOWERgmatRichC · Answer

Hi All,

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.

Final Answer:  D

GMAT assassins aren't born, they're made,
Rich

mvictor · Answer

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.

D

Mahmud6 · Answer

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

Bunuel · Answer

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

exc4libur · Answer

even-1=15, even=16 (possible)
even-1=18, even=19; or even+3=18, even=15 (not possible)
even-1=27, even=28 (possible)

Ans (D)

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Which of the following could be true of at least some of the terms of

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Which of the following could be true of at least some of the terms of

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