Events & Promotions
|
|
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.
Apr 3001:30 AM EDT
-02:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2912:30 AM EDT
-01:30 AM EDT
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
May 0306:30 AM PDT
-08:30 AM PDT
Verbal trouble on GMAT? Fix it NOW! Join Sunita Singhvi for a focused webinar on actionable strategies to boost your Verbal score and take your performance to the next level.
19 Mar 2024, 06:59
Kudos
Bookmarks
If \(x\) is an integer, how many solutions exist for the equation \(|x| + |x + 1| = |x|!\)?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
19 Mar 2024, 06:59
Kudos
Bookmarks
Official Solution:
If \(x\) is an integer, how many solutions exist for the equation \(|x| + |x + 1| = |x|!\)?
A. 0
B. 1
C. 2
D. 3
E. 4
The equation may appear daunting at first glance, yet, a closer examination reveals a simpler path to solution. Notice that the left-hand side invariably results in an odd number (when \(x\) is odd, \(x + 1\) becomes even, and their sum is odd; similarly, if \(x\) is even, then \(x + 1\) is odd, resulting in an odd sum). On the other side, the right-hand side represents the factorial of a non-negative integer, which is only odd for the cases of \(0! = 1\) and \(1! = 1\). For any other number, the factorial becomes even, as it includes a factor of 2. Consequently, we only need to consider those values of \(x\) for which \(|x|! = odd = 1\), specifically, -1, 0, and 1.
Substituting these values into the equation shows that \(x=0\) and \(x=-1\) are the only valid roots. Thus, the given equation has two solutions.
Answer: C
If \(x\) is an integer, how many solutions exist for the equation \(|x| + |x + 1| = |x|!\)?
A. 0
B. 1
C. 2
D. 3
E. 4
The equation may appear daunting at first glance, yet, a closer examination reveals a simpler path to solution. Notice that the left-hand side invariably results in an odd number (when \(x\) is odd, \(x + 1\) becomes even, and their sum is odd; similarly, if \(x\) is even, then \(x + 1\) is odd, resulting in an odd sum). On the other side, the right-hand side represents the factorial of a non-negative integer, which is only odd for the cases of \(0! = 1\) and \(1! = 1\). For any other number, the factorial becomes even, as it includes a factor of 2. Consequently, we only need to consider those values of \(x\) for which \(|x|! = odd = 1\), specifically, -1, 0, and 1.
Substituting these values into the equation shows that \(x=0\) and \(x=-1\) are the only valid roots. Thus, the given equation has two solutions.
Answer: C
15 May 2024, 08:51
Kudos
Bookmarks
any other alternative explanation?
