Last visit was: 12 Nov 2024, 16:24 It is currently 12 Nov 2024, 16:24
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

GRE Prep 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 GRE 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.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

The quantity will end in how many zeros

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 29944
Own Kudos [?]: 36204 [8]
Given Kudos: 25902
Send PM
The quantity will end in how many zeros [#permalink] New post   09 Aug 2017, 15:41
1
Expert Reply
7
Bookmarks
00:00

Question Stats:

66% (01:42) correct 33% (02:01) wrong based on 227 sessions

Hide Show timer Statistics



The quantity \(3^3 4^4 5^5 6^6\) - \(3^6 4^5 5^4 6^3\) will end in how many zeros ?

A. 3

B. 4

C. 5

D. 6

E. 9
ShowHide Answer
Official Answer
B
Most Helpful Community Reply
User avatar
B
amorphous
Retired Moderator
Joined: 07 Jan 2018
Posts: 739
Own Kudos [?]: 1442 [6]
Given Kudos: 93
Send PM

Top Member of the Month
Re: The quantity will end in how many zeros [#permalink] New post   08 Feb 2018, 22:51
6
realize that for a \(0\) to occur there has to be a multiplication of \(5\) and \(2\)

simplify the 1st term \(3^3 4^4 5^5 6^6\)= \(3^3* (2^2)^4*5^5* (2*3)^6\) = \(3^9* 2^1^4* 5^5\)
Similarly simplify the 2nd term that should come out to be \(3^9* 2^1^3* 5^4\)
Subtracting 2nd term from 1st term:take the common term which is whole of the 2nd term

\(3^9*2^1^3*5^4\)\((10-1)\)
now we have to find out the number of zeros in the common term because non common term is 9
2 and 5 multiply to 10. Here the limiting number is 5 which is equal to 4 hence 4 zeros
General Discussion
avatar
YMAkib
Intern
Intern
Joined: 08 Dec 2017
Posts: 40
Own Kudos [?]: 69 [0]
Given Kudos: 0
Send PM
Re: The quantity will end in how many zeros [#permalink] New post   07 Feb 2018, 01:18
1
3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3
=...(2^14)(5^5) - ....(2^13)(5^4)
There should be 4 zeroes ending up this quantity.
The answer is B.
avatar
gremather
Manager
Manager
Joined: 15 Feb 2018
Posts: 53
Own Kudos [?]: 34 [0]
Given Kudos: 0
Send PM
Re: The quantity will end in how many zeros [#permalink] New post   23 Feb 2018, 05:21
Hi, I don't quite understand your last sentence. What do you mean by the limiting number?

amorphous wrote:
realize that for a \(0\) to occur there has to be a multiplication of \(5\) and \(2\)

simplify the 1st term \(3^3 4^4 5^5 6^6\)= \(3^3* (2^2)^4*5^5* (2*3)^6\) = \(3^9* 2^1^4* 5^5\)
Similarly simplify the 2nd term that should come out to be \(3^9* 2^1^3* 5^4\)
Subtracting 2nd term from 1st term:take the common term which is whole of the 2nd term

\(3^9*2^1^3*5^4\)\((10-1)\)
now we have to find out the number of zeros in the common term because non common term is 9
2 and 5 multiply to 10. Here the limiting number is 5 which is equal to 4 hence 4 zeros
User avatar
B
amorphous
Retired Moderator
Joined: 07 Jan 2018
Posts: 739
Own Kudos [?]: 1442 [0]
Given Kudos: 93
Send PM

Top Member of the Month
Re: The quantity will end in how many zeros [#permalink] New post   26 Feb 2018, 09:34
2
gremather wrote:
Hi, I don't quite understand your last sentence. What do you mean by the limiting number?

amorphous wrote:
realize that for a \(0\) to occur there has to be a multiplication of \(5\) and \(2\)

simplify the 1st term \(3^3 4^4 5^5 6^6\)= \(3^3* (2^2)^4*5^5* (2*3)^6\) = \(3^9* 2^1^4* 5^5\)
Similarly simplify the 2nd term that should come out to be \(3^9* 2^1^3* 5^4\)
Subtracting 2nd term from 1st term:take the common term which is whole of the 2nd term

\(3^9*2^1^3*5^4\)\((10-1)\)
now we have to find out the number of zeros in the common term because non common term is 9
2 and 5 multiply to 10. Here the limiting number is 5 which is equal to 4 hence 4 zeros


since for a '\(0\)' to occur \(5\) has to be multiplied by \(2\). The number of zeros will depend on the minimum power raised of either of the two numbers

For eg.
\(100 = 5^2 * 2^2 = 2\) zeros at the end (because both the terms have power raised to 2)
\(125 = 5^3 * 2^0 = 0\) zeros at the end (because 2 is raised to a power of 0 hence 2 becomes the limiting number)
avatar
bellavarghese
Manager
Manager
Joined: 18 Jun 2019
Posts: 122
Own Kudos [?]: 42 [0]
Given Kudos: 0
Send PM
Re: The quantity will end in how many zeros [#permalink] New post   06 Sep 2019, 07:34
Hi! looking for another solution to this question.

I've simplified upto 3^9*2^13*5^4(9) and don't know how to proceed to fnd the number of zero's.

@greenlighttestprep, @Carcass - please help!
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 29944
Own Kudos [?]: 36204 [0]
Given Kudos: 25902
Send PM
Re: The quantity will end in how many zeros [#permalink] New post   06 Sep 2019, 08:04
1
Expert Reply
The trick is how many "five you do have in the quantity ??

we do have nine five numbers.

Now, you will have a zero whenever you do have \(2 \times 5 = 10\)

In our quantity we do have 4 couples of 5 plus one 5 disparaged

5*5
5*5
5*5
5*5
5

That means two couples of five are 4 zeros.

Hope this helps
avatar
bellavarghese
Manager
Manager
Joined: 18 Jun 2019
Posts: 122
Own Kudos [?]: 42 [0]
Given Kudos: 0
Send PM
Re: The quantity will end in how many zeros [#permalink] New post   06 Sep 2019, 08:07
Carcass wrote:
The trick is how many "five you do have in the quantity ??

we do have nine five numbers.

Now, you will have a zero whenever you do have 2 \times 5 = 10

In our quantity we do have 4 couples of 5 plus one 5 disparaged

5*5
5*5
5*5
5*5
5

That means two couples of five are 4 zeros.

Hope this helps



Much clearer!! Thank you!
User avatar
huda
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 708 [1]
Given Kudos: 161
Send PM

Top Member of the Month
Re: The quantity will end in how many zeros [#permalink] New post   09 Dec 2019, 02:56
1
Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



The quantity \(3^3 4^4 5^5 6^6\) - \(3^6 4^5 5^4 6^3\) will end in how many zeros ?

A. 3

B. 4

C. 5

D. 6

E. 9


Still if someone needs to be more clear visit here.
https://math.stackexchange.com/questions/2663286/this-quantity-see-question-will-end-with-how-many-zeros
avatar
hongj77
Intern
Intern
Joined: 15 Nov 2020
Posts: 9
Own Kudos [?]: 14 [3]
Given Kudos: 0
Send PM
Re: The quantity will end in how many zeros [#permalink] New post   26 Nov 2020, 12:29
3

More clearly explaining "limiting number"



Let's say you ended up with \(3^{19}2^{13}5^{4}\) and you want to know how many 0's it ends in.

Solution: Just factor out 10s



\(3^{19}2^{13}5^{4} = 3^{19}2^{9}(2*5)^{4} = 3^{19}2^{9}(10)^{4}.\)

\(10^{4}\) tells you there's 4 zeros.
avatar
DFXR
Intern
Intern
Joined: 25 Oct 2020
Posts: 24
Own Kudos [?]: 10 [1]
Given Kudos: 0
Send PM
Re: The quantity will end in how many zeros [#permalink] New post   03 Dec 2020, 23:22
1
I do not at all understand how the concept of limiting numbers applies in this case. Yes, the first term has 5 zeroes; the second one has 4. I don't see how subtraction of one from the other then results in 4, and being told "well it's the limiting number" doesn't really illuminate that much.

I can see how it works mathematically, in, say, the case of 100,000-10,000 but I am hesitant to apply this to other situations more broadly with confidence.
avatar
DylanGillo
Intern
Intern
Joined: 03 Dec 2020
Posts: 5
Own Kudos [?]: 1 [0]
Given Kudos: 0
Send PM
Re: The quantity will end in how many zeros [#permalink] New post   04 Dec 2020, 04:09
I think the right variant is 'C', or am I mistaken?
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 29944
Own Kudos [?]: 36204 [0]
Given Kudos: 25902
Send PM
Re: The quantity will end in how many zeros [#permalink] New post   04 Dec 2020, 05:25
Expert Reply
DylanGillo wrote:
I think the right variant is 'C', or am I mistaken?


what do you mean for the exact "variant" Sir '?
User avatar
bumpbot
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5008
Own Kudos [?]: 74 [0]
Given Kudos: 0
Send PM
Re: The quantity will end in how many zeros [#permalink] New post   14 Sep 2023, 23:00
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
gmatclubot
Re: The quantity will end in how many zeros [#permalink]
14 Sep 2023, 23:00
Moderators:
adewale223
GRE Instructor
78 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1111 posts
VibhuAnurag
GRE Instructor
234 posts

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

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact