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.
May 1408:30 AM PDT
-09:30 AM PDT
Most GMAT test-takers are intimidated by the hardest GMAT Verbal questions. In this session, Target Test Prep GMAT instructor Erika Tyler-John, a 100th percentile GMAT scorer, will show you how top scorers break down challenging Verbal questions..
May 1212:00 PM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.- May 13
08:30 AM PDT
-09:30 AM PDT
In Episode 4 of our GMAT Ninja CR series, we tackle the most intimidating CR question type: Boldface & "Legalese" questions. If you've ever stared at an answer choice that reads, "The first is a consideration introduced to counter a position that... - Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
16 Sep 2014, 00:47
Kudos
Bookmarks
What is the value of \(1! + 2! + ... + 10!\)?
A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918
A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918
16 Sep 2014, 00:47
Kudos
Bookmarks
Official Solution:
What is the value of \(1! + 2! + ... + 10!\)?
A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918
For any integer \(n\) greater than or equal to 5, the units digit of \(n!\) is zero. This is because, for \(n \geq 5\), \(n!\) will contain at least one 2 and at least one 5 as factors. When multiplied together, these produce a trailing zero. Hence, the terms from \(5!\) through \(10!\) each have zero as their units digit. Adding up the remaining terms, we'd have \(1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33\). Therefore, we see that the entire sum will have a units digit of 3 + 0 = 3. Only option B provides a number with 3 as its units digit.
Alternatively, we can do the following:
\(1! + 2! + ... + 10! = 1 + (2! + ... + 10!)\) = odd + even = odd.
\(1! + 2! + ... + 10! = 3 + (3! + ... + 10!)\) = integer divisible by 3.
Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.
Answer: B
What is the value of \(1! + 2! + ... + 10!\)?
A. 4,037,910
B. 4,037,913
C. 4,037,915
D. 4,037,916
E. 4,037,918
For any integer \(n\) greater than or equal to 5, the units digit of \(n!\) is zero. This is because, for \(n \geq 5\), \(n!\) will contain at least one 2 and at least one 5 as factors. When multiplied together, these produce a trailing zero. Hence, the terms from \(5!\) through \(10!\) each have zero as their units digit. Adding up the remaining terms, we'd have \(1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33\). Therefore, we see that the entire sum will have a units digit of 3 + 0 = 3. Only option B provides a number with 3 as its units digit.
Alternatively, we can do the following:
\(1! + 2! + ... + 10! = 1 + (2! + ... + 10!)\) = odd + even = odd.
\(1! + 2! + ... + 10! = 3 + (3! + ... + 10!)\) = integer divisible by 3.
Among the listed choices we are looking for odd integers divisible by 3. Only choice B fits.
Answer: B
General Discussion
15 Oct 2023, 08:19
Kudos
Bookmarks
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
