GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 18 Jun 2019, 18:07

Close

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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

How many consecutive zeros will appear at the end of 43! if that numbe

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Manager
Manager
avatar
Joined: 23 Sep 2016
Posts: 93
How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

Show Tags

New post 02 Dec 2016, 14:09
4
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

68% (00:41) correct 32% (01:25) wrong based on 138 sessions

HideShow timer Statistics

How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?

A) 12
B) 11
C) 10
D) 9
E) 8
Senior Manager
Senior Manager
avatar
B
Joined: 13 Oct 2016
Posts: 364
GPA: 3.98
Re: How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

Show Tags

New post 02 Dec 2016, 23:14
\(\frac{43}{5} + \frac{43}{5^2} = 8 + 1 = 9\)

Answer D
Intern
Intern
avatar
Joined: 18 Sep 2016
Posts: 2
GMAT ToolKit User
Re: How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

Show Tags

New post 04 Dec 2016, 09:18
SW4 wrote:
How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?

A) 12
B) 11
C) 10
D) 9
E) 8


Thanks to user Bunuel, unforunately first post so cannot link, but formula copied below:
\(\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n

For number of zeroes denominator must be less than 43, therefore use:
-> 5 as 5 < 43
-> 5^2 as 25 < 43
-> 5^3 cannot as 125 >43

=\(\frac{43}{5} + \frac{43}{5^2}\)
= 8 + 1
=9

Ans D
Board of Directors
User avatar
P
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4504
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)
GMAT ToolKit User
Re: How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

Show Tags

New post 04 Dec 2016, 10:39
SW4 wrote:
How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?

A) 12
B) 11
C) 10
D) 9
E) 8


43/5 = 8
8/5 = 1

Hence, no of zeroes will be 9

Answer will be (D)


_________________
Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55670
Re: How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

Show Tags

New post 05 Dec 2016, 01:40
Target Test Prep Representative
User avatar
D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6567
Location: United States (CA)
Re: How many consecutive zeros will appear at the end of 43! if that numbe  [#permalink]

Show Tags

New post 14 Sep 2018, 17:42
SW4 wrote:
How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?

A) 12
B) 11
C) 10
D) 9
E) 8


To determine the number of trailing zeros in a number, we need to determine the number of 5- and-2 pairs within the prime factorization of that number. Each 5-and-2 pair creates a 10, and each 10 creates an additional zero.

Since we know there are fewer 5s in 43! than 2s, 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 43!, we can use the following shortcut in which we divide 43 by 5, then divide the quotient of 43/5 by 5 and continue this process until we no longer get a nonzero quotient.

43/5 = 8 (we can ignore the remainder)

8/5 = 1 (we can ignore the remainder)

Since 1/5 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 43!.

Thus, there are 8 + 1 = 9 factors of 5 within 43! This means we have nine 5-and-2 pairs, so there are 9 consecutive zeros at the end of 43! when it is expanded to its final result.

Answer: D
_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

GMAT Club Bot
Re: How many consecutive zeros will appear at the end of 43! if that numbe   [#permalink] 14 Sep 2018, 17:42
Display posts from previous: Sort by

How many consecutive zeros will appear at the end of 43! if that numbe

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne