It is currently 14 Dec 2017, 15:02

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at

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

Hide Tags

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42607

Kudos [?]: 135660 [1], given: 12705

30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

Show Tags

New post 12 Nov 2017, 00:48
1
This post received
KUDOS
Expert's post
4
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

38% (01:10) correct 62% (01:22) wrong based on 60 sessions

HideShow timer Statistics

\(30^{30} * 29^{29} * 28^{28}*...*2^2 * 1^1 = n\). How many zeroes does ‘n’ contain at the end of the number (to the right of the last non-zero digit)?

A. 30
B. 60
C. 63
D. 105
E. 130
[Reveal] Spoiler: OA

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135660 [1], given: 12705

2 KUDOS received
Manager
Manager
avatar
Joined: 17 Mar 2015
Posts: 121

Kudos [?]: 57 [2], given: 4

Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

Show Tags

New post 12 Nov 2017, 01:54
2
This post received
KUDOS
3
This post was
BOOKMARKED
The original expression can be rewritten as \(\frac{30!^{30}}{29!*28!*27!*...*3!*2!*1!}\).
Using the forumla for factorials and the trailing zeroes:
\(\frac{30}{5} + \frac{30}{25} = 7\)
7 trailing zeroes for 30!, so the numerator has 210 zeroes.
Denominator has, using the same approach:
6 for 29!, 28!, 27!, 26! and 25!
4 for 24!, 23!, 22!, 21! and 20!
3 for 19!, 18!, 17!, 16! and 15!
2 for 14!, 13!, 12!, 11! and 10!
1 for 9!, 8!, 7!, 6! and 5!
0 for 4!, 3!, 2! and 1!.
Product of numbers with trailing zeroes makes these zeroes add up, then the resulting tally of zeroes in the denominator is: \(5*(6+4+3+2+1) = 5*16 = 80\)
Division with trailing zeroes makes you subtract the total of trailing zeroes in the denominator from the total of trailign zeroes in the numerator
\(210 - 80 =130\)

E

Kudos [?]: 57 [2], given: 4

3 KUDOS received
Intern
Intern
avatar
Joined: 17 Oct 2017
Posts: 3

Kudos [?]: 3 [3], given: 0

Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

Show Tags

New post 12 Nov 2017, 02:22
3
This post received
KUDOS
Count the number of 5's to get the number of Zeroes
Only 5,10,15,20,25,30 contains 5's
30^30 has 30 5's
25^25 has 50 5's
20^20 has 20 5's
15^15 has 15 5's
10^10 has 10 5's
5^5 has 5 5's
So the total is 130


Sent from my SM-G925I using GMAT Club Forum mobile app

Kudos [?]: 3 [3], given: 0

Expert Post
5 KUDOS received
Math Expert
User avatar
D
Joined: 02 Aug 2009
Posts: 5349

Kudos [?]: 6128 [5], given: 121

Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

Show Tags

New post 12 Nov 2017, 03:33
5
This post received
KUDOS
Expert's post
3
This post was
BOOKMARKED
Bunuel wrote:
\(30^{30} * 29^{29} * 28^{28}*...*2^2 * 1^1 = n\). How many zeroes does ‘n’ contain at the end of the number (to the right of the last non-zero digit)?

A. 30
B. 60
C. 63
D. 105
E. 130


Number of 10s will depend on number of 2s and 5s
Here since 5 is the larger prime number, it will give us the number of 0s
So number involving 5 are \(30^{30}*25^{25}*20^{20}*15^{15}*10^{10}*5^5=5^{30}*(5^2)^{25}*5^{20}*5^{15}*5^{10}*5^5=5^{30+2*25+20+15+10+5}=5^{130}\)
Ans 130
E
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6128 [5], given: 121

Intern
Intern
avatar
B
Joined: 03 Jul 2017
Posts: 5

Kudos [?]: 5 [0], given: 0

Location: India
Concentration: Finance, Accounting
WE: Information Technology (Computer Software)
Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

Show Tags

New post 12 Nov 2017, 04:32
To find the no. of zeros find either
1) no. of 5's in the series( as 5*2 =10 makes the zero)
2) if a no. contains the zeros itself (like 30, 20 and 10 in above series)

In the above series we have enough two's, all even no(28,26,24....2) will provide us the two. We have to only look for the 5's and that will come from 5, 15,25.
in 5^5 gives 5 5's
in 15^5 gives 15 5's
in 25^5 gives 50 5's

therefore total- 70 5's i.e. 70 zeros.

also, if a no. contain a zero itself when multiply itself it will have the no. of zeros equals to( no. of zeros in the no. at the right end)* (power of that no).

30^30 will give 30 zeros.
20^20 will give 20 zeros
10^10 will give 10 zeros

Total- 60 zeros


Overall we have then 60+70= 130 zeros.

Also see the attachment for more detail


Please hit the kudos button if you like this :)
Attachments

IMG_2.jpg
IMG_2.jpg [ 1.55 MiB | Viewed 565 times ]

IMG_1.jpg
IMG_1.jpg [ 1.72 MiB | Viewed 563 times ]

Kudos [?]: 5 [0], given: 0

Manager
Manager
avatar
B
Joined: 11 Feb 2017
Posts: 175

Kudos [?]: 10 [0], given: 198

Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

Show Tags

New post 13 Nov 2017, 06:20
Bunuel wrote:
\(30^{30} * 29^{29} * 28^{28}*...*2^2 * 1^1 = n\). How many zeroes does ‘n’ contain at the end of the number (to the right of the last non-zero digit)?

A. 30
B. 60
C. 63
D. 105
E. 130



Can someone explain this question in most easiest way???

Kudos [?]: 10 [0], given: 198

Intern
Intern
avatar
B
Joined: 03 Jul 2017
Posts: 5

Kudos [?]: 5 [0], given: 0

Location: India
Concentration: Finance, Accounting
WE: Information Technology (Computer Software)
Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

Show Tags

New post 13 Nov 2017, 11:18
rocko911 wrote:
Bunuel wrote:
\(30^{30} * 29^{29} * 28^{28}*...*2^2 * 1^1 = n\). How many zeroes does ‘n’ contain at the end of the number (to the right of the last non-zero digit)?

A. 30
B. 60
C. 63
D. 105
E. 130



Can someone explain this question in most easiest way???

Question is asking how many zeros will come at the right end if we mutiply the above series.


Suppose if we multiple : 101*100 then 10100 will be d answer and u can see before any digit other than zero appear we already have two zero at the right end .... So answer for this is two zeros....



Similarly for the above series we have to find how many continous zeros will come at right side before any other digit will appear ....


For d logic you can see the attached image ...

Just remember if we multiply any no. Than zero in the end will come if either the no. Has zero at the end in it like 10, 20, 300,. 400,....


Or if a the multiplying no. Has factor of 10 i.e one 5's and one 2's .... As 5*2 gives us 10


Eg 75 *8 ( in 75 we have 2 5's ( 3*5*5) and in 8 we have 3 tow's ) but we have only 2 fives and 3 twos so no. Of zero will come in the end is two)


75*8 = 600...

See above attachment and hit the kudos if u like this :) thanks

Sent from my Redmi Note 4 using GMAT Club Forum mobile app

Kudos [?]: 5 [0], given: 0

Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at   [#permalink] 13 Nov 2017, 11:18
Display posts from previous: Sort by

30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at

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


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.