|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
31 Dec 2023, 10:40
Kudos
Bookmarks
If \(n\) is a positive two-digit even integer, which of the following must be a factor of \((n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5)\)?
I. 11
II. 21
III. 45
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
I. 11
II. 21
III. 45
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
31 Dec 2023, 10:40
Kudos
Bookmarks
Official Solution:
If \(n\) is a positive two-digit even integer, which of the following must be a factor of \((n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5)\)?
I. 11
II. 21
III. 45
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
Since \(n\) is a positive two-digit even integer, then \((n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5)\) is a product of six consecutive odd integers. For consecutive odd integers:
• One of every three is a multiple of 3;
• One of every five is a multiple of 5;
• One of every seven is a multiple of 7;
• One of every nine is a multiple of 9;
• One of every eleven is a multiple of 11;
And so on.
Hence, out of six consecutive odd integers, we for sure will have two multiples of 3 and at least one multiple of 5. Thus, \((n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5)\) will always be divisible by \(3*3*5 = 45\).
However, we cannot guarantee divisibility by either 11 or 7 (thus by 21).
For example, if the six consecutive odd integers are 13, 15, 17, 19, 21, and 23, their product is not divisible by 11, because there is no multiple of 11 among them. Similarly, if the six consecutive odd integers are 9, 11, 13, 15, 17, and 19, their product is not divisible by 7, because there is no multiple of 7 among them.
Answer: C
If \(n\) is a positive two-digit even integer, which of the following must be a factor of \((n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5)\)?
I. 11
II. 21
III. 45
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
Since \(n\) is a positive two-digit even integer, then \((n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5)\) is a product of six consecutive odd integers. For consecutive odd integers:
• One of every three is a multiple of 3;
• One of every five is a multiple of 5;
• One of every seven is a multiple of 7;
• One of every nine is a multiple of 9;
• One of every eleven is a multiple of 11;
And so on.
Hence, out of six consecutive odd integers, we for sure will have two multiples of 3 and at least one multiple of 5. Thus, \((n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5)\) will always be divisible by \(3*3*5 = 45\).
However, we cannot guarantee divisibility by either 11 or 7 (thus by 21).
For example, if the six consecutive odd integers are 13, 15, 17, 19, 21, and 23, their product is not divisible by 11, because there is no multiple of 11 among them. Similarly, if the six consecutive odd integers are 9, 11, 13, 15, 17, and 19, their product is not divisible by 7, because there is no multiple of 7 among them.
Answer: C
