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

[AAPL]

Magoosh

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

OA D.

[fskilnik@GMATH]

Magoosh

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

From the fact that n is the index in the sequence, we know (implicitly) that n is a positive integer.

I. May the expression (product) given have (at least) one 3 AND one 5?
Sure. Take n=3, for instance, so that b_3 = 5*9 and we have our needs satisfied!

II. May the expression (product) given have (at least) one 2 AND two 3´s?
NO. Reason: (2n-1) and (2n+3) are both ODD numbers (for any integer value of n).

III. May the expression (product) given have (at least) three 3´s?
Sure. Take n=14, because (2n-1) = 27 , therefore (2n-1)*(2n+3) = 27*(positive integer) ...

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.

[Scott@TargetTestPrep]

Magoosh

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

OA D.

We see that no matter what n is, (2n - 1) and (2n + 3) will each be odd. So b_n is odd, and it can never be divisible by 18, which is even. Now, if one of the factors of b_n is 15 (for example, if n = 8, we have 2n - 1 = 15), then b_n will be divisible by 15. Similarly, if one of the factors of b_n is 27 (for example, if n = 12, we have 2n + 3 = 27), then b_n will be divisible by 27. Therefore, we see that b_n can be divisible by 15 and 27 for some values of n, but it can't be divisible by 18 for any values of n.

Answer: D