• NEW! FREE Beat The GMAT Quizzes
Hundreds of Questions Highly Detailed Reporting Expert Explanations
• 7 CATs FREE!
If you earn 100 Forum Points

Engage in the Beat The GMAT forums to earn
100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## How many trailing Zeroes does 53! + 54! have? tagged by: VJesus12 ##### This topic has 3 expert replies and 0 member replies ## How many trailing Zeroes does 53! + 54! have? How many trailing Zeroes does 53! + 54! have? (A) 12 (B) 13 (C) 14 (D) 15 (E) 16 The OA is B. Experts, how can I know it without computing the sum? ### GMAT/MBA Expert GMAT Instructor Joined 22 Aug 2016 Posted: 1975 messages Followed by: 30 members Upvotes: 470 VJesus12 wrote: How many trailing Zeroes does 53! + 54! have? (A) 12 (B) 13 (C) 14 (D) 15 (E) 16 The OA is B. Experts, how can I know it without computing the sum? A trailing zero to any number occurs when the number is multiplied by 10. Since 10 has two prime factors: 2 and 5, we must count how many 2s and 5s are there in a number. Since 2 < 5, it is clear that the number of 2s â‰¥ number of 5s. Thus, we must only count the number of 5s. We have 53! + 54!. 53! + 54! = 53!(1 + 54) = 53!.55 = 53!.5.11 Let's count the number of 5s in 53!.5.11. The number of 5s in 53! = 1.2.3.... 53 will be available in 5, 10, 15, 20, 25, 30, 35, 40, 45, & 50. There would be two 5s each in 25 and in 50. Thus, the total number of 5s in 53! = 12. We have one more 5 in 53!.5.11. Thus, the total number of 5s in 53!.5.11 = 13. The correct answer: B Hope this helps! -Jay Download free ebook: Manhattan Review GMAT Quantitative Question Bank Guide _________________ Manhattan Review GMAT Prep Locations: New York | Hyderabad | Mexico City | Toronto | and many more... Schedule your free consultation with an experienced GMAT Prep Advisor! Click here. ### GMAT/MBA Expert Legendary Member Joined 14 Jan 2015 Posted: 2667 messages Followed by: 122 members Upvotes: 1153 GMAT Score: 770 I'd use the exact same logic Jay employed here. Here's a neat shortcut you can use, once you know you're looking for the number of 5's in the prime factorization of 53! * 5 * 11: Number of terms in 53! with at least one five in its prime factorization:: 53/5 = 10 + 3/5, so there are 10 terms with at least one five. Number of terms in 53! with at least two fives in its prime factorization:: 53/(5^2) = 2 + 3/25, so there are 2 terms with at least two fives. (Because 53/(5^3) < 1, there are no terms that have three fives.) Number of 5's in 53!: 10 + 2 = 12 If there are twelve 5's in 53! and one more five in the produce 53! * 5 * 11, there will be a total of 12 + 1 = 13 fives. _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now!

### GMAT/MBA Expert

GMAT Instructor
Joined
09 Apr 2015
Posted:
1465 messages
Followed by:
18 members
39
VJesus12 wrote:
How many trailing Zeroes does 53! + 54! have?

(A) 12
(B) 13
(C) 14
(D) 15
(E) 16
To determine the number of trailing zeros, we need to determine the number of 5-and-2 pairs within the prime factorization of 53! + 54!. Letâ€™s start by simplifying 53! + 54!.

53! + 54! = 53!(1 + 54) = 53!(55)

Since we know there are fewer 5s than 2s in 53!(55), we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

To determine the number of 5s within 53!, we can use the following shortcut in which we divide 53 by 5, then divide the quotient of 53/5 by 5 and continue this process until we no longer get a nonzero quotient.

53/5 = 10 (we can ignore the remainder)

10/5 = 2

Since 2/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 53!.

Thus, there are 10 + 2 = 12 factors of 5 within 53!.

Finally, we see that there is one factor of 5 within 55.

Since there are 13 factors of 5 within 53!(55), there are thirteen 5-and-2 pairs and thus 13 trailing zeros.

_________________

Jeffrey Miller
jeff@targettestprep.com

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

• Free Trial & Practice Exam
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

• Free Veritas GMAT Class
Experience Lesson 1 Live Free

Available with Beat the GMAT members only code

• Award-winning private GMAT tutoring
Register now and save up to $200 Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

Available with Beat the GMAT members only code

• Free Practice Test & Review
How would you score if you took the GMAT

Available with Beat the GMAT members only code

• 5 Day FREE Trial
Study Smarter, Not Harder

Available with Beat the GMAT members only code

• Get 300+ Practice Questions

Available with Beat the GMAT members only code

• 5-Day Free Trial
5-day free, full-access trial TTP Quant

Available with Beat the GMAT members only code

• 1 Hour Free
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

• Magoosh
Study with Magoosh GMAT prep

Available with Beat the GMAT members only code

### Top First Responders*

1 Ian Stewart 57 first replies
2 Brent@GMATPrepNow 31 first replies
3 Jay@ManhattanReview 29 first replies
4 GMATGuruNY 21 first replies
5 ceilidh.erickson 15 first replies
* Only counts replies to topics started in last 30 days
See More Top Beat The GMAT Members

### Most Active Experts

1 Scott@TargetTestPrep

Target Test Prep

199 posts
2 Max@Math Revolution

Math Revolution

84 posts
3 Brent@GMATPrepNow

GMAT Prep Now Teacher

69 posts
4 Ian Stewart

GMATiX Teacher

65 posts
5 GMATGuruNY

The Princeton Review Teacher

40 posts
See More Top Beat The GMAT Experts